AND 

THEIR  EQUATIONS 


MEMCAL 


COLLEGE  OF  PHARMACY 


O&Nfemla  Oo!!e£©  of  Pharmacy 


CHEMICAL  REACTIONS 
AND  THEIR  EQUATIONS 

HACKH 


CHEMICAL   REACTIONS 
AND  THEIR  EQUATIONS 


A  Guide  and  Reference  Book 
for  Students  of  Chemistry 


BY 

INGO  W.  D.  HACKH,  PH.C.,  A.B. 

PROFESSOR   OF  BIOCHEMISTRY,    COLLEGE  OF   PHYSICIANS 


AND   SURGEONS  OF   SAN   FRANCISCO 


Q&Ilfernta,  College  of  Pharmaos 


PHILADELPHIA 
P.    BLAKISTON'S  SON   &   CO. 

1012  WALNUT   STREET 


COPYRIGHT,  1921,  BY  P.  BLAKISTON'S  SON  &  Co. 


THE  MAPLE  PRESS  YORK  PA 


PREFACE 

The  inability  to  balance  a  chemical  equation  is  a  most 
common  difficulty  of  students  of  chemistry.  The  writer 
when  teaching  at  a  large  university  actually  encountered 
graduate  students  of  chemistry  who  were  unable  to 
balance  an  incomplete  ionic  equation  that  involved 
oxidation  and  reduction,  not  to  mention  the  large 
number  of  first  and  second  year  students  who  had  only 
a  very  hazy  idea  of  the  principles. 

In  order  to  supply  students  with  necessary  material 
and  to  expound  the  general  principles  of  balancing  equa- 
tions, this  concise  volume  was  written.  It  does  not  enter 
into  a  detailed  discussion  of  physico-chemical  equations, 
but  confines  itself  mainly  to  a  consideration  of  purely 
chemical  equations  from  a  technical  and  arithmetical 
standpoint.  The  writer  hopes  that  a  study  of  this 
volume,  (which,  to  the  writer's  knowledge,  is  the  only 
one  of  its  kind  on  the  subject)  in  connection  with  any 
good  text,  will  aid  toward  a  clearer  understanding  of 
chemistry  on  the  part  of  the  student.  A  detailed  study 
of  the  following  pages  should  enable  anyone  to  balance 
any  chemical  equation  rapidly  and  correctly. 

With  pleasure  the  writer  acknowledges  his  gratitude 
to  Dr.  Henry  D :  Arcy  Power  for  many  valuable  sugges- 
tions and  criticisms  in  reading  the  manuscript,  and  to 
his  students,  Messrs.  Jordan,  Konrad,  and  Magee  for 
reading  the  manuscript  and  proofs.  He  also  feels 
much  indebted  to  his  teachers,  Profs.  Edmond  O'Neill, 
G.  N.  Lewis,  and  Joel  H.  Hildebrand. 

I.W.D.H. 

SAN  FRANCISCO,  CAL. 

v 

4O  d">  * }  (I 
6060 


TABLE  OF  CONTENTS 

CHAPTER  I 

PAGE 

SYMBOLS 1 

Atoms,  molecules,  ions,  ionization. 

CHAPTER  II 

FORMULAS 6 

Empirical,  rational,  constitutional,  and  structural  formulas. 
Valency  and  valence  numbers.     Oxidation  and  reduction. 
Nomenclature  and  terminology  of  compounds. 
Summary  of  information  contained  in  a  formula. 

CHAPTER  III 

EQUATIONS  (INVOLVING  NO  OXIDATION  AND  REDUCTION)         ...     23 
Molecular  and  ionic  equations.     Finishing  and  balancing  equa- 
tions.    Calculations  and  problems. 

CHAPTER  IV 

EQUATIONS  (INVOLVING  OXIDATION  AND  REDUCTION) 37 

Molecular  and  ionic  equations.     The  three  steps  in  balancing. 
Examples  and  problems. 

CHAPTER  V 

REACTIONS  AND  THEIR  CONTROL 48 

The  speed  of  chemical  reactions  and  the  chemical  equilibrium. 
Mechanical  control  by  surface,  catalysts,  concentration,  and 

pressure.     Thermal  control  and  the  influence  of  heat. 
Electrical  control,  electromotive  force,  galvanic  cells,  electrolysis. 

CHAPTER  VI 

TYPES  OF  CHEMICAL  REACTIONS  AND  EQUATIONS 78 

Analysis,  synthesis,  and  metathesis.  Neutralization  and 
hydrolysis.  Combination,  dissociation,  displacement,  and 
substitution. 


viii  TABLE    OF   CONTENTS 

APPENDIX 

PAGE 
I.  KEY  TO  NOMENCLATURE  WITH  A  LIST  OF  RADICALS,  IONS,  AND 

VALENCE  NUMBERS 103 

II.  DISPLACEMENT  SERIES  OF  THE  ELEMENTS 107 

III.  PERIODIC  SYSTEM  AND  CLASSIFICATION  OF  THE  ELEMENTS  .    .    .  109 

IV.  SOLUBILITY  TABLE  OF  COMPOUNDS 115 

V.  PREPARATION  OF  SALTS 118 

KEY  TO  THE  EQUATIONS      119 

INDEX  AND  GLOSSARY  OF  CHEMICAL  TERMS 133 


ERRATA 


Page  3:  line  13.    Place  comma  after  "therefore"  and  take  out  comma 

after  "atom." 

Page  101,  equation  No.  429.    Read  Zn++  instead  of  Zn     . 
Page  112,  in  table  of  atomic  weight  read  175.0  instead  of  17. 50  for  Lutecium. 
Page  112,  line  8,  read  an  instead  of  n. 

English  prices : 

Page  47,  question  27,  last  sentence  would  read:  Fe  costs  3d,  Al  =  sh.  1.3d, 

and  Mg  =  sh.  1.6d. 
Page  45,  question  2.    Price  of  Zn  =  2  sh. 


HACKH,  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS. 


CHEMICAL  REACTIONS  AND  THEIR 
EQUATIONS 

CHAPTER  I 
SYMBOLS 

Symbols. — The  alphabet  of  chemistry  consists  of. 
symbols.  Every  chemical  element  is  represented  by 
a  symbol  which  is  usually  the  first  letter  of  its  English 
or  Latin  name.  Where  different  elements  have  the 
same  initial  letter  a  small  additional  letter  is  added, 
e.g.  B  =  boron,  Be  =  beryllium,  Br  =  bromine,  Bv  = 
brevium.  A  symbol  not  only  represents  the  kind  of 
element  but  also  stands  for  a  certain  quantity  of  it; 
namely  the  smallest  amount  of  matter  that  can  enter 
into  combination — a  single  atom.  This  amount  or 
quantity  is  expressed  in  a  relative  number — the  atomic 
weight,  which,  as  such,  is  a  number  indicating  the 
proportional  weight  of  an  element  in  combination. 

Atoms  and  Molecules. — We  assume  that  atoms  can 
only  exist  when  combined  in  the  form  of  a  molecule; 
thus  atoms  of  like  nature  combine  to  form  an  elemental 
molecule  and  aggregate  to  an  element,  while  unlike 
atoms  combine  to  form  a  compound  molecule  and 
aggregate  to  a  compound.  Molecules  are  capable  of 
independent  existence,  that  is,  they  may  exist  separately 
and  not  in  aggregation  with  similar  molecules.  Thus 
a  molecule  is  the  smallest  imaginable  particle  of  a 
compound,  an  atom  the  smallest  imaginable  particle 
of  an  element.  While  atoms  are  chemically  inde- 

l 


2  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

structible,1  the  molecules  are  susceptible  to  endless 
chemical  change.  Such  changes  and  interchanges  are 
expressed  in  the  form  of  chemical  equations. 

A  chemical  symbol  has  three  distinct  uses  as  it  may 
represent  an  atom,  -molecule  or  ion.  To  distinguish 
whether  a  symbol  represents  an  atom,  molecule,  or  ion 
a  numeral  or  +  or  —  sign  is  used. 

Atom. — The  atom  of  an  element  is  shown  by  the 
symbol  alone.  Thus  H  represents  an  atom  of  hydrogen 
with  an  atomic  weight  of  1.008;  Cl  stands  for  an  atom 
of  chlorine  with  an  atomic  weight  of  35.45,  and  Fe  is 
an  atom  of  iron  (ferrum)  with  atomic  weight  56.5. 

Molecule. — The  molecule  of  an  element  is  shown  by 
the  symbol  with  a  small  numeral  attached  indicating  the 
number  of  atoms:  H2  or  H2  is  a  molecule  of  hydrogen 
containing  two  atoms  of  hydrogen,  the  molecular  weight 
being  the  sum  of  the  atomic  weight  =  2  X  1.008  = 
21016;  S4  or  S4  is  a  molecule  of  sulfur  containing  four 
atoms  of  sulfur,  the  molecular  weight  being  4  X  32.00 
=  128.00;  S6  or  S6  is  a  molecule  of  sulfur  with  six  atoms 
of  sulfur,  the  molecular  weight  of  this  molecule  is 
6  X  32.00  =  192.00;  Sx  or  S*  is  a  molecule  of  sulfur 
containing  an  unknown  number  of  atoms  of  sulfur. 
All  gaseous  elements  have  ordinarily  two  atoms  in  their 
molecule,  e.g.  O2,  N2,  C12,  which  indicates  respectively 
a  molecule  of  oxygen,  nitrogen,  and  chlorine  as  gases. 

1  The  fact  that  atoms  are  chemically  indestructible  does  not  exclude 
the  possibility  that  they  may  be  transformed  by  physical  means.  Mod- 
ern research,  both  theoretical  and  practical,  point  to  still  smaller  cor- 
puscles or  electrons  as  the  building  stones  of  atoms.  Nevertheless 
the  conception  of  a  "structure  of  the  atoms"  will  not  affect  the  older 
conception  of  an  atom  as  the  "chemical  unit,"  or  an  element  as  a  type  of 
atoms.  Chemists  will  thus  continue  to  deal  with  atoms,  and  future 
progress  in  the  subatomic  realm  will  merely  increase  the  knowledge  of 
the  relation  among  elements  and  their  probable  evolution,  but  will  not 
materially  change  the  application  of  the  atomic  theory. 


SYMBOLS  3 

Only  in  a  few  cases  have  the  number  of  atoms  in  a 
molecule  been  determined  for  solid  elements,  e.g.  P4, 
P8,  As2,  As4.  If  the  number  of  atoms  in  a  molecule 
of  a  solid  element  is  unknown,  the  molecule  should  be 
written  with  x,  e.g.  Nax,  Fex,  AU,  Cx.  However,  in 
metals,  this  number  is  assumed  to  be  one,  so  that  in 
this  case  Na  would  represent  not  only  an  atom,  but  also 
a  molecule  of  sodium  (natrium). 

Ion. — The  ion  of  an  element  is  indicated  by  the 
symbol  with  one  or  more  positive(+)or  negative  (  — ) 
signs  attached,  which  represent  the  number  of  positive 
or  negative  charges  attached  to  the  atom  of  an  element. 
An  ion,  therefore  is  an  electrified  atom,  or  an  atom  with 
a  certain  number  of  electrical  corpuscles  attached. 
These  charges  are  also  written  as  dots  and  dashes,  thus  : 
H+  or  H*  =  hydrogen  ion,  H-atom  with  one  positive 
charge,  Cl~  or  Cl'  =  chloride  ion,  Cl-atom  with  one 
negative  charge,  Ca++  or  Ca"  =  calcium  ion,  Ca-atom 
with  two  positive  charges,  S~'~  or  S"  =  sulfide  ion, 
S-atom  with  two  negative  charges,  A1+++  or  Al'"  = 
aluminum  ion,  Al-atom  with  three  positive  charges. 
These  ions  ordinarily  exist  only  in  solutions.  The 
majority  of  salts  dissolved  in  water  become  ionized 
because  their  component  atoms  become  electrified. 
Thus  NaCl  (sodium  chloride)  when  dissolved  in  water 
will  form  Na+  (sodium  ion)  and  Gl~  (chloride  ion), 
while  a  small  amount  of  the  original  sodium  chloride 
remains  as  NaCl  (molecular  or  non-ionized  sodium 
chloride).  Some  salts  will  ionize  slightly,  that  is,  they 
will  break  apart  (dissociate)  only  to  a  small  degree. 
Ammonium  chloride  (NH4C1)  when  dissolved  in  water 
forms  a  few  ammonium  ions  (NH4+)  and  a  few 
chloride  ions  (Cl.~) — the  larger  percentage  of  the  sub- 
stance remaining  in  molecular  combination  (NH4C1). 


4  CHEMICAL   REACTIONS    AND    THEIR    EQUATIONS 

lonization  or  electric  dissociation,  therefore,  is  the 
breaking  apart  of  the  molecule  in  solution  whereby 
the  atoms  or  groups  of  atoms  (radicals)  acquire  a 
positive  or  negative  charge.  It  must  be  borne  in 
mind  that  the  properties  of  the  ions  are  distinctly 
different  from  the  properties  of  the  atoms.  Sodium  is 
different  from  sodium  ion,  iron  metal  (Fe)  differs  from 
ferrous  ion  (Fe++)  and  ferric  ion  (Fe+++),  for  the  ions 
are  always  electrified  or  charged  atoms. 

Summary. — The  three  ways  of  using  a  chemical  symbol 
are  summarized  as  follows: 

H  =  one  hydrogen  atom,  which  exists  in  compounds, 
2H  =  two  hydrogen  atoms, 

H2  =  one  hydrogen  molecule,  which  exists  in  hydro- 
gen gas, 

2H2  =  two  hydrogen  molecules, 
H+  =  one  hydrogen  ion,  which  exists  in  the  solution 

of  acids, 
2H+  =  two  hydrogen  ions. 

Accordingly  the  symbols  Cl,  S,  28,  38,  Fe,  Al,  stand 
respectively  for  one  atom  of  chlorine,  one  atom  of  sulfur, 
two  atoms  of  sulfur,  three  atoms  of  sulfur,  one  atom 
of  iron,  one  atom  of  aluminum,  while  their  respective 
atomic  weights;  C12,  SB,  2S6,  3S6,  Fe«,  Alx,  means  one 
molecule  of  chlorine  (two  atoms),  one  molecule  of  sulfur 
(six  atoms),  two  molecules  of  sulfur  (twelve  atoms), 
three  molecules  of  sulfur  (eighteen  atoms),  one  molecule 
of  iron  (unknown  number  of  atoms),  one  molecule  of 
aluminum  (unknown  number  of  atoms);  Cl~,  8  , 
28—,  38—,  Fe++,  Fe+++,  A1+++,  represent  respectively 
one  chloride  ion  (chlorine  atom  with  one  negative 
charge),  one  sulfide  ion  (sulfur  atom  with  two  negative 
charges),  two  sulfide  ions  (two  sulfur  atoms  with  four 


SYMBOLS  O 

negative  charges),  three  sulfide  ions  (three  sulfur  atoms 
with  six  negative  charges),  one  ferrous  ion  (one  iron  atom 
with  two  positive  charges),  one  ferric  ion  (one  iron  atom 
with  three  positive  charges),  one  aluminum  ion  (one 
aluminum  atom  with  three  positive  charges). 

QUESTIONS 

1.  Define  symbol,  atom,  molecule,  ion,  ionization. 

2.  What  is  the  symbol  for  an  atom  of  bromine,  selenium,  potassium, 
nitrogen,  copper,  magnesium,  phosphorus? 

3.  What  is  the  symbol  for  a  molecule  of  the  elements  given  in  question 
2? 

4.  What  is  the  symbol  for  an  ion  of  the  elements  given  in  question  2? 
6.  What  is  the  difference  between  N2  and  2N,  O2  and  O3,  3O2,  2O3  and 

6O,  F2,  2F-and2F? 

6.  State    all    the  information  contained  in  the  symbols:  Na,    Na+, 
Se,  Se4,  Se— ,  C,  As,  As2,  As4,  Hg,  Hg+,  Hg++. 


CHAPTER  II 
FORMULAS 

Formula. — The  vocabulary  of  chemistry  consists  of 
formulas.  A  chemical  formula  is  a  combination  of 
symbols  which  shows  the  number  and  kind  of  atoms 
in  a  molecule,  hence,  the  formula  tells  the  chemical 
composition  of  a  substance  and  represents  the  molecule 
of  a  compound.  However,  as  the  symbols  not  only 
indicate  the  kind  of  atom,  but  also  a  certain  quantity, 
called  the  atomic  weight,  it  is  evident  that  a  formula 
must  also  include  the  quantitative  composition  of  a 
compound,  and  the  molecular  weight. 

For  instance,  NaCl  means  a  molecule  of  sodium 
chloride  containing  one  atom  of  sodium  or  23  parts  by 
weight,  and  one  atom  of  chlorine  or  35.5  parts  by  weight. 
The  total  or  molecular  weight  is  naturally  the  sum  of 
the  atomic  weights,  that  is,  23  +  35.5  =  58.5.  From 
this  proportion  the  percentage  can  be  calculated,  for 
as  58.5  parts  of  sodium  chloride  contain  23  parts  of 
sodium,  then  58.5  :  23  :  :  100  :  x,  and  x  =  23  X  100  /58.5 
=  39.3  that  is  39.3  per  cent  sodium;  while  the  per- 
centage of  chlorine  is  found  by  the  proportion  58.5  : 
35.5  :  :  100  :  x,  and  x  =  35.5  X  100  /  58.5  =  60.7  which 
means  60.7  per  cent  of  chlorine.  Two  molecules  of 
sodium  chloride  may  be  written  either  2NaCl  or  (NaCl)  2- 
In  the  first  case  the  two  in  front  refefs  to  the  whole 
following  formula,  in  the  second  case  the  parenthesis 
is  necessary  as,  otherwise,  the  formula  NaCl2  would 
indicate  a  compound  molecule  containing  one  atom  of 
sodium  and  two  atoms  of  chlorine  which  is  wrong. 

6 


FORMULAS 


Na2SO4  means  one  molecule  of  sodium  sulfate  contain- 
ing two  atoms  of  sodium  or  2  X  23  =  46  parts  of  sodium, 
one  atom  of  sulfur  or  32  parts  of  sulfur, 

four  atoms  of  oxygen  or  4  X  16       =64  parts  of  oxygen, 


which  added  together  make  142  parts  of  sodium 

sulfate  and  the  molecular  weight  therefore  is  142.0. 

Na2SO4.7H20  means  a  molecule  of  crystallized  sodium 
sulfate  in  which  there  are  in  addition  to  the  sodium, 
sulfur,  and  oxygen,  seven  molecules  of  water.  This 
water  which  forms  an  essential  part  of  the  crystal  is 
called  the  water  of  crystallization.  The  water  of  crystal- 
lization can  be  driven  off  by  heat  and  the  substance 
becomes  anhydrous. 

The  formula  for  ammonium  aluminum  sulfate, 
A12(SO4)3.(NH4)2SO4.24H2O,  indicates  a  more  complex 
compound,  a  molecule  of  which  contains  one  molecule 
of  aluminum  sulfate  (A12 (804)3),  one  molecule  of  am- 
monium sulfate  ((NH4)2SO4),  and  twenty-four  molecules 
of  water.  Adding  these  molecules  together,  the  number 
of  individual  atoms  in  this  compound  are: 

2  atoms  of  aluminum  or  2  X  27.1     =  54.2  parts  of  Al 

3  +  1=4  atoms  of  sulfur  or  4  X  32  =  128.0  parts  of  S 
2  atoms  of  nitrogen  or  2  X  14           =  28 . 0  parts  of  N 
(3  X  4)  +  4  +  24  =  40  atoms  of 

oxygen  or  40  X  16   =640.0  parts  of  O 
(2X4)  +  (24  X  2)  =  56  atoms 

of  hydrogen  or  56  X  1    =    56 . 0  parts  of  H 


making  altogether  906 . 2  parts  of 

crystallized  ammonium  aluminum  sulfate,   having    the 
molecular  weight  of  906.2. 


8  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

Types  of  Formulas. — The  formula  for  this  compound 
may  also  be  written  Al2S4N204oH56.  Such  a  formula  is 
called  an  empirical  formula  for  only  the  number  and  kind 
of  atoms  are  given.  From  such  a  formula  the  quantita- 
tive composition  of  the  substance  can  be  calculated,  but 
it  does  not  indicate  the  way  in  which  the  atoms  are 
arranged  in  the  molecule.  The  arrangement  of  the 
atoms,  that  is,  the  way  in  which  they  group  themselves, 
is  shown  by  the  rational  formula.  The  rational  formula 
given  above  for  ammonium  alum  shows  that  sulfur  is 
present  in  the  sulfate  radical  (S04),  and  nitrogen  is 
present  in  the  ammonium  radical  (NH4),  while  some  of 
the  oxygen  and  hydrogen  is  present  as  water  of  crystalliza- 
tion. Thus  the  rational  formula  here  shows  (a)  two 
atoms  of  aluminum  are  linked  to  three  sulfate  radicals, 
(6)  two  ammonium  radicals  are  linked  to  one  sulfate 
radical,  and  (c)  the  crystals  contain  twenty-four  mole- 
cules of  water. 

There  are  four  types  of  formulas:  The  empirical,  the 
rational,  the  constitutional  and  the  structural  or  graph- 
ical. The  characteristics  of  each  type  are  as  follows: 

Empirical  Formula. — The  empirical  formula  shows 
only  the  number  and  kind  of  atoms  from  which  (to 
repeat)  the  molecular  weight  and  the  percentage  com- 
position can  be  calculated.  Thus  F2S3Oi2  simply  means 
a  molecule  containing  two  atoms  of  iron,  three  atoms 
of  sulfur,  and  twelve  atoms  of  oxygen,  the  molecular 
weight  being  (2  X  55.9)  +  (3  X  32)  +  ( 12 X  16)  =  399.8. 
Such  a  type  of  formula  gives  nothing  further  regarding 
the  chemical  nature  and  is  practically  never  used  in 
inorganic  chemistry,  they  are  used,  however,  in  organic 
chemistry  for  compounds  of  which  only  the  percentage 
composition  is  known,  and  which  have  not  Been  fully 
investigated. 


FORMULAS  9 

Rational  Formula.  —  The  rational  formula  shows  not 
only  the  number  and  kind  of  atoms,  but  also  indicates 
the  way  in  which  the  atoms  are  linked  together.  Thus 
Fe2(SO4)3  means  that  two  atoms  of  iron  are  linked  to 
three  sulfate  radicals,  each  sulfate  radical  containing 
one  atom  of  sulfur  and  four  atoms  of  oxygen.  Such  a 
formula  gives  some  information  regarding  the  chemical 
characteristics  of  the  compound,  e.g.  this  compound 
would  give  the  characteristic  sulfate  reaction  (a  precipi- 
tate with  barium  chloride  solution),  etc. 

Constitutional  Formula.  —  The  constitutional  formula 
is  a  notation  which  throws  further  light  upon  the  in- 


tramolecular  arrangement  of  the  atoms.     Thus 


illustrates  the  valency  bonds  and  shows  that  two  triva- 
lent  iron  atoms  are  linked  to  three  bivalent  SO4  groups, 
in  other  words  it  shows  that  the  iron  in  this  compound  has 
three  bonds,  and  that  the  sulfate  radical  has  two  bonds. 
Structural  Formula.  —  The  structural  formula  is  the  most 
highly  developed  form  of  notation  and  illustrates  the  link- 
age of  each  individual  atom,  every  bond  being  represented 


by   a  line.     Thus        n/^^O  indicates  that  the  iron 
Fe^0\c/0 


is  not  directly  linked  to  the  sulfur  atom,  as  may  be  as- 
sumed from  the  constitutional  formula,  but  is  connected 
by  an  oxygen  atom  to  the  sulfur.  It  also  shows  that 
there  are  six  single  bonds  between  two  iron  and  six  oxy- 
gen atoms,  likewise  six  single  bonds  between  three  sulfur 


10  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

and  six  oxygen  atoms,  and  finally  six  double  bonds 
between  three  sulfur  and  six  oxygen  atoms.  In  a 
structural  formula  the  radicals  are  resolved  into  their 
simplest  components—  single  atoms,  and  the  connection 
between  each  atom  is  shown.  Thus  the  above  structure 
formula  indicates  that  oxygen  is  always  bivalent,  as 
it  has  two  bonds,  while  the  sulfur  is  hexavalent  having 
six  bonds. 

In  inorganic  chemistry  the  rational  formula  is  gen- 
erally used,  while  the  constitutional  and  structural 
formulas  together  with  the  empirical  formula  are  mainly 
employed  in  organic  chemistry.  In  organic  chemistry 
the  structure  theory  has  yielded  the  most  magnificent 
results,  for  without  the  recognition  of  a  structure,  none 
of  the  synthetic  substances  could  be  manufactured  such 
as  indigo,  camphor,  dyes  tuffs,  and  pharmaceuticals. 
A  modern  simplification  of  structure  formulas  for  organic 
compounds  has  been  devised  in  the  form  of  structure 
symbols  (see  Canadian  Chemical  Journal,  vol.  2,  p.  135, 
1918;  Science,  vol.  48,  p.  333,  1918;  Chem.  News,  vol. 
118,  p.  289,  1919). 

Summary.  —  Summarized,  the  four  types  of  formula  are: 

Empirical  :  A12C309  C3H4O4 

Rational  :  A12  (CO3)  3         CH2  (CO2H)  2 

Constitutional:  »          Ab=CO3  /COOH 

CH/ 
X 


Structural  or  graphic  :        /O\rx 


=  0 


H 

O 


FORMULAS  11 

Valency. — What  is  a  bond  or  valency?  How  is  it 
possible  to  depict  graphically  the  intramolecular  arrange- 
ment of  the  atoms?  Logically  we  reason  that  the 
number  of  atoms  in  a  molecule  must  depend  upon  the 
force  which  holds  the  atoms  together;  furthermore  we 
find  that  certain  groups  of  atoms,  such  as  the  sulfate 
radical,  are  very  stable  and  pass  through  a  number  of 
chemical  changes  as  units,  and  in  such  cases  the  force 
holding  the  atoms  together  must  be  especially  strong. 
At  present  we  have  no  means  of  measuring  this  force 
nor  do  we  know  its  mechanism.  However,  one  of  its 
manifestations  is  that  property  of  the  element  which 
is  termed  valency.  Valency  is  sometimes  defined  as  the 
combining  power  of  an  atom  in  regard  to  hydrogen,  but 
it  is  more  exact  to  define  valency  as  the  combining 
capacity  of  an  atom  or  group  of  atoms  (radicals)  using 
an  atom  of  hydrogen  as  a  unit  of  comparison.  Hence 
if  an  atom  of  element  X  combines  with  one  atom  of 
H  and  forms  a  compound  HX,  then  X  is  said  to  have 
a  valency  of  one.  If  the  atom  of  element  Y  can  combine 
with  two  atoms  of '  hydrogen  to  form  a  compound  of 
the  type  H2Y,  then  the  valency  of  Y  is  two,  and  like- 
wise if  an  atom  of  element  Z  combines  with  three  atoms 
of  hydrogen  to  give  a  molecule  of  the  type  H3Z,  then 
Z  has  a  valency  of  three.  The  structural  formulas 

H\  H\ 

for  these   three  molecules  are   H-X,       YY,  and  H-)Z 

W 

in  which  each  valency  is  shown  by  a  line.  In  a  com- 
pound the  atoms  which  have  a  valency  of  one  are 
called  monovalent  (e.g.  H,  Na,  K,  Cl,  etc.),  of  two  = 
bivalent  (e.g.  O,  Ca,  Ba,  etc.),  of  three  =  trivalent 
(e.g.  Al),  of  four  =  tetravalent  (e.g.  C,  Si,  etc.),  of  five  = 


12  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

pentavalent,  of  six  =  hexavalent,  of  seven  =  heptavalent, 
and  of  eight  =  octovalent. 

Valence  Number. — It  has  become  necessary  to  divide 
the  elements  into  negative  and  positive  groups  (see 
Chapter  5)  and  accordingly  in  each  compound  there  is 
a  negative  and  positive  constituent.  This  assumption 
leads  to  the  consideration  of  positive  and  negative  valence 
numbers  (also  called  polar  numbers).  The  basis  for  the 
assignment  of  valence  numbers  is  hydrogen  with  a  positive 
valence  number  of  one  or  +1.1  In  all  stable  compounds 
the  sum  of  the  valence  numbers  must  equal  zero  and 
by  considering  H  always  as  +1  the  positive  or  nega- 
tive character  of  the  other  elements  can  be  established. 
Thus  in  HX  the  atom  X  must  have  a  valence  number  of 

—  1,  in  H2Y  the  atom  Y  has  a  valence  number  of  —  2,  in 
H3Z  the  atom  Z  has  a  valence  number  of  —3.     Therefore 
it  follows  that  Cl,  in  HC1,  must  have  a  valence  number  of 

—  1,  that  O,  in  H20,  has  a  valence  number  of  —  2,  and  N, 
in  NH3,  has  a  valence  number  of  —3. 

In  some  cases  elements  do  not  combine  with  hydrogen, 
but  replace  hydrogen  (see  displacement  series),  because 
they  are  more  positive  than  hydrogen,  and  accordingly 
have  a  positive  valence  number.  Thus  Na,  which 
replaces  one  atom  of  H,  has  a  valence  number  of  +1; 
calcium  replaces  two  atoms  of  hydrogen  and  has  a 
valency  of  two  and  valence  number  of  +2.  The 
positive  valence  numbers  range  from  +1  to  +8,  while 
the  negative  valence  numbers  range  from  —1  to  —4. 
The  elements  in  their  free  state  (uncombined)  are 
assumed  to  have  a  valence  number  of  0. 

Univalent  and  Polyvalent. — However,  it  is  a  common 
misapprehension  to  assume  that  a  given  element  can 

1  Only  in  a  very  few  cases  has  H  a  valence  number  of  —  1,  such  com- 
pounds are  the  hydrides,  e.g.  NaH,  LiH. 


FORMULAS  13 

form  only  one  type  of  compound,  that  is  can  have  only 
one  definite  valency  or  valence  number  for  in  many 
cases  an  element  can  form  two  or  more  series  of  com- 
pounds, that  is,  have  two  or  more  definite  valencies  or 
valence  numbers.  The  first  class  of  elements  are  spoken 
of  as  univalent  elements  (which  must  not  be  confused 
with  monovalent  elements),  while  the  second  class  of 
elements  are  called  polyvalent  elements.  In  general 
the  alkali  metals  (Na,  K,  etc.),  earthalkali  metals 
(Ca,  Ba,  etc.),  and  earth  metals  (Al,  etc.)  are  univalent, 
while  the  non  metals  and  heavy  metals  are  polyvalent. 

A  univalent  element  may  consist  of  monovalent  atoms 
(e.g.  Na),  bivalent  atoms  (e.g.  Ca),  trivalent  atoms 
(e.g.  Al),  and  so  on.  A  polyvalent  element  may  consist 
of  mono-  and  bivalent  atoms  (Cu),  mono-  and  trivalent 
atoms  (Au),  bi-  and  trivalent  atoms  (Fe),  bi-,  tri- 
and  hexavalent  atoms  (Cr),  mono-,  tri-,  penta-,  and 
hexavalent  atoms  (Cl),  and  so  on. 

Series  of  Compounds. — A  polyvalent  element,  that 
is,  an  element  having  two  or  more  valencies,  will  form 
two  or  more  series  of  compounds  and  these  series  of 
compounds  will  have  distinct  properties  and  different 
names.  Thus  iron  has  three  series  of  compounds, 
namely  ferrous  compounds  such  as  FeCl2,  FeO,  etc.  in 
which  the  valency  is  two  and  the  valence  number 
+2;  ferric  compounds  such  as  FeCl3,  Fe2O3,  etc.  in 
which  the  valency  of  iron  is  three  and  the  valence  number 
+3;  and  finally  the  ferrates  such  as  Na2FeO4  in  which 
the  valency  of  iron  is  six  and  the  valence  number  +6. 
Chlorine  has  four  series  of  compounds :  the  chlorides 
(NaCl,  BaCl2,  etc.)  in  which  chlorine  has  a  valency  of 
one  and  a  valence  number  of  —  1 ;  the  chlorites  (NaOCl, 
Ca(OCl)2)  in  which  chlorine  has  a  valency  of  three, 
and  a  valence  number  of  +3;  the  chlorates  (KC103) 


14  CHEMICAL   REACTIONS   AND   THEIR   EQUATIONS 

with  a  valency  of  five  and  valence  number  +5;  and 
finally  the  perchlorates  (KC1O4)  with  a  valency  of  seven 
and  valence  number  of  +7.  Though  chlorine  is  a  nega- 
tive element,  like  all  non  metals,  it  has  in  only  one  series 
of  compounds,  the  chlorides,  a  negative  valence  number, 
and  in  the  other  three  series  a  positive  valence  number. 
This  positive  character  of  the  valence  number  for  a 
negative  element  is  caused  by  assuming  the  valence 
number  of  oxygen  in  a  compound  to  be  always  —2. 
This  conception  aids  in  understanding  oxidation  and 
reduction,  as  explained  further  on. 

Finding  the  Valence  Number. — From  the  formula  of 
a  compound  the  valence  number  of  the  atoms  and  their 
positive  or  negative  character  can  readily  be  found  by 
remembering  that  hydrogen  in  a  compound  is  always 
+  1,  and  oxygen  in  a  compound  is  always  —  2,  and  that 
the  sum  of  the  valence  numbers  in  a  saturated  and 
stable  compound  must  always  be  zero.  Thus,  in  water 
(H2O)  there  is  H2  =  2  X  (  +  1)  =  +2,  and  O  =  -2, 
therefore  (+2)  +  (-2)  =  0.  In  lime  (CaO)  oxygen 
is  —2,  so  calcium  must  be  +2.  In  ferric  oxide  (Fe2O3) 
there  are  three  oxygen  atoms,  that  is  3 (  —  2)  =  (  —  6),  so 
the  two  iron  atoms  must  be  +6,  and  one  iron  atom  +3, 
and  the  valence  number  of  iron  in  that  compound  is  +3. 
In  a  similar  way  the  valence  numbers  for  all  other 
atoms  are  determined  and  can  readily  be  found  by  the 
difference  necessary  to  make  the  sum  zero,  hence^it  is 
simply  an  arithmetical  problem.  A  few  examples  will 
illustrate  this. 

Potassium  permanganate  has  the  formula  KMnO4. 
To  find  the  valence  number  for  the  Mn  atom  add  the 
valence  number  of  K  (  +  1)  to  the  sum  of  the  valence 
numbers  of  oxygen  4  (—2)  =  —8  and  get  —  7,  thus  Mn 
must  have  +7,  To  make  this  clearer  write  the  known 


FORMULAS  15 

valence  numbers  under  the  symbols  and  form  an 
arithmetical  equation  thus: 

K      Mn        O4 

(  +  1)  +  x  +  4(-2)  =  0  and  hence  x  =  +7. 

The  valence  number  of  nitrogen  in  sodium  nitrate  is 

Na        N         03 

(  +  1)  +  x  +  3(-2)  =  0.  In  this  case  x  =  +5,  thus 
the  valence  number  of  nitrogen  is  +5.  Likewise  the 
valence  number  of  chromium  in  potassium  bichromate 

is  found:  K2         Cr2  OT 

2(  +  l)  +  2x  +  7(-2)  =  0,  from  this  follows 
that  2x  =  12  and  the  valence  number  of  one  chromium 
atom  is  +6. 

Oxidation  and  Reduction. — If  an  element  has  several 
valence  numbers,  that  is,  if  it  forms  several  series  of 
compounds,  we  speak  of  different  stages  of  oxidation. 
As  a  rule  oxidation  can  be  defined  as  the  increase  or 
augmentation  of  the  valence  number,  that  is,  a  change 
from  a  lower  to  a  higher  valence  number;  while  reduction 
is  the  decrease  or  diminution  of  the  valence  number, 
that  is,  the  change  from  a  higher  to  a  lower  valence 
number.  The  transformation  of  ferrous  compounds 
(+2)  into  ferric  compounds  (+3)  is  thus  oxidation, 
while  the  transformation  of  a  ferrous  compound  (+2) 
into  metallic  iron  (0)  is  reduction.  In  the  former  the 
ferrous  compound  has  been  oxidized,  in  the  latter  it  has 
been  reduced.  Again  if  a  chloride  (  —  1)  is  transformed 
into  chlorine  gas  (0),  the  valence  number  has  been 
increased  and  the  chlorine  atom  in  the  chloride  is  oxi- 
dized to  free  chlorine.  If  chlorine  gas  (0)  is  further 
oxidized  to  a  chlorite  (+3),  chlorate  (+5),  or  perchlorate 


16  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

(+7),  the  valence  number  is  further  increased  and 
therefore  the  chlorine  atom  further  oxidized.  Reduction 
is  a  change  in  the  opposite  direction,  e.g.  the  chlorine 
atom  in  a  perchlorate  (+7)  can  be  reduced  to  chlorate, 
chlorite,  chlorine,  or  chloride.  Schematically  the  case 
is  illustrated  by  the  following: 

Valence  numbers:  -1  0  +3  +5  +7 

Type  of  compounds :  chlorides  chlorine  chlorites  chlorates  perchloratea 
Formula:  XC1  C12  XC1O2  XC1O3  XC1O4 

OXIDATION > 

« REDUCTION 

in  which  X  is  a  monovalent  element. 

Any  change  from  left  to  right  is  oxidation. 

Any  change  from  right  to  left  is  reduction. 

It  is  evident  from  the  diagram  that  the  chlorides  can 
not  be  further  reduced,  that  the  perchlorates  can  not 
be  further  oxidized,  and  that  free  chlorine  gas  can  be 
either  reduced  to  chlorides,  or  oxidized  to  chlorites, 
chlorates,  or  perchlorates. 

Similarly  there  are  several  series  of  sulfur  compounds : 

Valence  numbers:  —  2  0  -f4  4-6 

Types  of  compounds:       sulfides          sulfur  sulfites          sulfates 

Formula:  X2S  S  X2SO3  X2SO4 

OXIDATION > 

< REDUCTION 

The  conclusions  drawn  from  this  diagram  are: 

Sulfides  can  be  oxidized  to  free  sulfur,  sulfites,  or 
sulfates. 

Sulfates  can  be  reduced  to  sulfites,  free  sulfur,  or 
sulfides. 

Free  sulfur  can  be  oxidized  <to  sulfites  and  sulfates,  or 
reduced  to  sulfides. 

Sulfites  can  be  oxidized  to  sulfates,  and  reduced  to  free 
sulfur  and  sulfides. 


FOKMULAS  17 

Sulfides  cannot  be  reduced,  and  sulfates  cannot  be 
oxidized. 

It  is  advantageous  for  the  student  to  construct  such 
tables  and  diagrams  for  other  elements  which  have 
several  series  of  compounds,  such  as  N,  Br,  As,  P,  Fe,  Cr, 
Mn,  etc.,  or  to  devise  a  single  table  in  which  the  more 
common  elements  are  arranged  in  the  order  of  their 
compounds. 

Nomenclature. — The  different  types  or  series  of  com- 
pounds of  an  element  have  different  names.  These 
names  are  formed  by  the  addition  of  the  suffixes  -ides, 
-ites,  -ates  to  a  negative  element,  and  -ous  or  -ic  to  a 
positive  element. 

Endings  of  Names. — Thus  there  are  for  negative  ele- 
ments or  radicals  the  suffixes: 

-ides  for  the  lowest  form  of  oxidation,  characterized  by 
the  absence  of  oxygen  and  a  negative  valence  number 
(nitrides,  hydrides,  phosphides,  carbides,  selenides,  tel- 
lurides,  arsenides,  etc.). 

-ites  for  the  lowest  form  of  oxidation  containing  few 
atoms  of  oxygen,  characterized  by  the  lowest  positive  va- 
lence number  (nitrites,  phosphites,  sulfites,  arsenites,  etc.). 

-ates  for  the  normal  form  of  oxidation  containing  the 
normal  amount  of  oxygen  atoms  and  characterized  by 
the  normal  positive  valence  number  (nitrates,  sul- 
fates, phosphates,  carbonates,  arsenates,  manganates, 
chromates,  ferrates,  etc.). 

per-  -ates  for  the  highest  form  of  oxidation  containing 
most  oxygen  atom's  and  having  the  highest  positive 
valence  number  (persulfates,  permanganates,  perchlor- 
ates,  etc.). 

For  positive  elements  the  suffixes  are: 

-ous  indicating  the  lower  form  of  oxidation,  that  is,  the 


18  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

lower  positive  valence  number  (ferrous,  arsenous,  mercur- 
ous,  cuprous,  nickelous,  plumbous,  etc.). 

-ic  indicating  the  higher  form  of  oxidation,  the  higher 
positive  valence  number  (ferric,  arsenic,  mercuric,  cupric, 
nickelic,  plumbic,  etc.). 

Key  to  Nomenclature. — A  list  of  the  more  common 
elements,  ions,  and  radicals  with  their  valence  numbers 
can  be  found  in  the  appendix,  and  this  key  to  the  nomen- 
clature of  chemical  compounds  should  enable  the  student 
to  write  the  correct  formula  for  any  compound,  or  to 
give  the  correct  name  for  any  formula  of  inorganic 
chemistry. 

Summary  of  Information  Contained  in  a  Formula.— 
From  any  chemical  formula  there  can  be  deduced: 

(a)  The  kind  of  elements  and  the  number  of  atoms 
which  constitute  the  molecule. 

(6)  The  weight  relation  of  these  elements  and  the 
molecular  weight  of  the  compound. 

(c)  The  percentage  of  weight  of  the  elements  compos- 
ing the  molecule. 

(d)  The  valency  of  the  elements  and,  therefore,  their 
stage  of  oxidation. 

Whenever  the  formula  refers  to  a  gas  or  gaseous  com- 
pound there  is,  furthermore,  expressed  by  a  formula: 

(e)  The  volume  relation  of  the  gaseous  constituents. 
According  to  Avogadro's  law  there  are,  under  the  same 
pressure  and  at  the  same  temperature,  in  the  same  volume 
of  any  gas  the  same  number  of  molecules,  so  it  follows 
that  in  1  liter  of  oxygen  there  are  the  same  number  of 
molecules  as  in  1  liter  of  hydrogen,  and  as  in  the  forma- 
tion of  water,  2  molecules  of  H2  react  with  1  molecule  of 
O2,  so  2  liters  of  H2  are  required  to  combine  with  1  liter 
of  O2. 

(/)  The  specific  gravity  or  density  of  the  gas.     Density 


FORMULAS  19 

can  be  expressed  in  three  ways  according  to  the  unit 
chosen.  (1)  Density  with  regard  to  oxygen  =  D(02) 
refers  to  oxygen  gas  for  comparison  and  its  molecular 
weight  O2  =  32  is  taken  as  a  unit,  hence  the  molecular 
weight  of  any  other  gas  is  also  its  density  or  D(02)  = 
Mol.  Wt.  For  H2  the  D(02)  is  2,  for  N2  the  D(02)  is  28, 
etc.  (2)  Density  with  regard  to  hydrogen  =  D(H2) 
refers  to  the  lightest  known  gas  as  unity.  In  this  case 
the  molecular  weight  of  the  gas  must  be  divided  by  the 
molecular  weight  of  hydrogen,  that  is  D(H2)  =  Mol.Wt./2. 
Hence  for  O2  the  D(H2)  is  16,  for  N2  the  D(H2)  is 
14.  (3)  Density  in  regard  to  air  as  unity  =  D(air). 
A  certain  volume  filled  with  32  grams  oxygen  will  weigh, 
when  filled  with  air,  28.95  grams,  hence  if  the  density  of 
air  is  taken  as  unity,  the  density  of  oxygen  =  D(air)  is 
32/28.95  -  1.1053  and  accordingly  the  D(air)  of  all 
other  gases  is  found  by  dividing  their  molecular  weight 
by  28.95,  that  is  D(air)  =  Mol.  Wt./28.95. 

(g)  The  weight  of  1  liter  of  gas.  The  volume  occupied 
by  a  mole  (grammolecule)  of  any  gas  is  22.4  liters.  Thus 
a  mole  of  H2,  pr  2  grams  of  hydrogen,  will  fill  22.4  liters, 
a  mole  of  oxygen,  or  32  grams  of  oxygen  gas,  will  also 
occupy  22.4  liters  under  standard  conditions  of  pressure 
and  temperature,  hence  the  simple  proportion  22.4  :  M 
: :  1  :  x,  and  x  =  M/22.4  will  give  the  weight  of  1  liter 
of  any  gas.  In  the  above  equation  M  represents  the 
molecular  weight  in  grams  (  =  mole),  and  x  the  weight 
of  1  liter  of  gas  in  grams. 

(h)  The  volume  occupied  by  1  gram  of  any  gas.  In 
this  case  the  above  proportion  becomes  M  :  22.4  : :  1  :  y, 
and  y  =  22.4/M,  for  if  M  grams  will  fill  22.4  liters,  then 
1  gram  will  occupy  22.4/M  liters. 

Example. — The  formula  NH3  for  ammonia  means: 

(a)  The  molecule  consists  of  one  atom  of  nitrogen  and 


20  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

three  atoms  of  hydrogen  making  altogether  four  atoms 
in  the  molecule. 

(fe)  The  molecular  weight  is  the  sum  of  the  atomic 
weights,  thus  14  parts  of  nitrogen  (by  weight — not  vol- 
ume) and  3X1  =  3  parts  of  hydrogen  give  17  parts  of 
ammonia — hence  the  molecular  weight  is  17. 

(c)  The  percentage  of  nitrogen  and  hydrogen  is  calcu- 
lated from  the  data  given^in  (&),  thus  in  17  parts  am- 
monia there  are   14  parts  of  nitrogen,  then  17  :  14  :: 
100  :  x,    hence    14  X  100/17  =  x,    and    x  =  82.25    per 
cent  nitrogen,  while  17  :  3  ::  100  :  y  and  y  =  3  X 100/17 
gives  for  y  the  percentage  of  hydrogen  as  17.75  per 
cent. 

(d)  The  valence  number  of  H  being  +1,  three  atoms 
of  H  are  thus  equal  to  +3,  therefore  the  valence  number 
of  nitrogen  must  be   —3.     From  the  table  in  the  ap- 
pendix we  see  that  —3  is  the  lowest  stage  of  oxidation 
for  nitrogen  and  it  is  evident  that  ammonia  can  not  be 
further  reduced,  but  can  be  oxidized  to  free  nitrogen, 
nitrites,  or  nitrates. 

(e)  Nitrogen  and  hydrogen  are  both  gases,  therefore 
one  volume  of  nitrogen  gas  and  three  volumes  of  hydro- 
gen gas  will  combine  and  form  ammonia. 

(/)  The  specific  gravity  or  density  of  ammonia  (1) 
compared  with  C>2  =  32  equals  the  molecular  weight 
which  is  D(02)  =  17  for  ammonia;  (2)  compared  with 
H2  =  1,  it  equals  17/2,  that  is  D(H2)  =  8.5;  (3)  com- 
pared with  air  =  1,  it  equals  17/28.95,  that  is  D(air)  = 
0.59.  Ammonia  is  thus  a  little  over  half  as  light  as  air 
and  oxygen,  but  8.5  times  as  heavy  as  hydrogen. 

(g)  If  22.4  liters  of  ammonia  weigh  17  grams,  then  1 
liter  will  weigh  17/22.4  =  0.76  gram. 

(h)  If  17  grams  of  ammonia  occupy  22.4  liters,  then  1 
gram  will  occupy  22.4/17  =  1.31  liters. 


FORMULAS  21 

QUESTIONS 

1.  Define  (a)  formula,  (b)  valency,  (c)  valence  number,  (d)  oxidation, 
(e)  reduction,  (/)  radicals,  (g)  monovalent  and  univalent  elements. 

2.  Write  out  all  the  information  contained  in  the  formulas  (a)  KI, 
(6)  AgN03,  (c)  H3P04,  (d)  K2HP04,  (e)  MgSO4.5H2O. 

3.  Write  out  all  the  information  contained  in  the  formulas  for  the 
following  gaseous  compounds  (a)  HC1,  (b)  N2O3,  (c)  N2Os,  (d)  CH4. 

4.  Calculate   the   molecular   weights    of    the   following    compounds: 
(a)  ferrous  sulfate  FeSO4,  (b)  ferric  sulfate  Fe2(SO4)3,  (c)  glucose  C6Hi2O6, 
(d)  sugar  Ci2H22On,  (e)  bismuth  nitrate  Bi(NO3)3,  (/)  potassium  alum 
KA1(SO4)3.12H2O,    (g)  nitroglycerin   C3H5(NO3)3,  (h)  hemoglobin  C726- 
Hn7iNi94O2i3S3. 

5.  Calculate  the   percentage   composition   of    (a)    mercuric   chloride 
HgCl2,    (b)   mercurous   chloride   HgCl,    (c)   manganese  dioxide   MnO2, 
(d)   copper  sulfate   CuSO4.5H2O,    (e)   sodium   carbonate   Na2CO3,    (/) 
crystallized  sodium   carbonate  Na2CO3.7H2O,  (g)  sugar  Ci2H22On,  (h) 
ammonium  nitrate  NH4NO3,  (i)  ammonium  sulfate  (NH4)2SO4,  (j)  silver 
chloride  AgCl,  (k)  silver  acetate  AgC2H3O2.NB.     In  (d)  and  (/)  calculate 
the  percentage  of  water,  not  hydrogen  and  oxygen. 

6.  What  is  the  rational  formula  for  such  a  compound  as 


(a)    NH4\  rn  (6)  Fe%  (c)  Nax 

NH,7C°3 


\  TT^^»1U4         f 

Fe./P°4  H/ 

7.  What  is  the  structural  formula  for  the  compounds  given  in  6? 

8.  Determine  the  valence  numbers  of  the  following  elements: 

(a)  sulfur  in  H2S,  Na2SO3,  Na2SO4,  FeS,  Na2S,  FeSO4,  CaS,  CaSO4,  SO2, 
SO3,  H2SO4;  (6)  nitrogen  in  NH3,  NH4OH,  N2O2,  N2O,  N2O3,  NO,  NO2, 
HNO2,  HNO3,  NH4C1,  KNO2,  NH4NO3;  (c)  iron  in  FeO,  FeCl3,  Fe(OH)2, 
Fe(OH)3,  FeCNS,  Fe(CN)2,  K4Fe(CN)6,  K3Fe(CN)6.  NOTE.  The 
cyan  radical  (CN~)  has  a  valence  number  of  —1  like  Cl~;  in  it  we 
assume  C  with  —  4,  and  N  with  -f3. 

9.  Give  an  example  of  the  lowest  and  highest  stage  of  oxidation  of 
(a)  sulfur,  (6)  chlorine,  (c)  nitrogen,  (d)  manganese,  (e)  iron,  (/)  chromium, 
(g)  mercury,  (h)  copper,  (i)  lead. 

10.  Write  four  or  more  formulas  of  different  compounds  for  each  series 
of  compounds  or  stages  of  oxidation  of  (a)  sulfur,  (b)  chlorine,  (c)  iron, 
(d)  nitrogen,  and  give  the  proper  names  to  these  compounds. 

11.  Write  out  the  rational  formulas  for  the  following  compounds: 
(a)  arsenous  chloride,  (c)  arsenic  chloride,  (b)  cerous  nitrate,  (d)  eerie 
sulfate,  (e)  potassium  manganate,  (/)  sodium  permanganate,  (g)  calcium 
chloride,   (h)  calcium  chlorite,   (i)  calcium  chlorate,   (j)  potassium  per- 


22  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

chlorate,  (k)  mercuric  bromate,  (I)  mercurous  arsenate,  (m)  mercuric 
arsenite,  (n)  ferrous  bromide,  (o)  ferric  bromide,  (p)  ferrous  bromate, 
(q)  ferric  bromate,  (r)  potassium  stannate,  (s)  sodium  stannite,  (t)  lithium 
plumbite,  (u)  potassium  plumbate,  (/>)  niccolous  carbonate,  (w)  niccolic 
carbonate,  (x)  plumbic  bichromate,  (y)  plumbous  chromate,  (z)  plumbic 
chromate.  NOTE.  Consult  the  key  to  nomenclature  in  the  appendix. 

12.  Which  of  the  following  ores  contains  the  highest  percentage  of  iron : 
(a)   hematite  Fe2O3,   (b)  magnetite  Fe3O4,   (c)  limonite  (Fe3O4)2.3H2O, 
(d)  siderite  FeCO3,  (e)  pyrite  FeS2? 

13.  Calculate  the  percentages  of  carbon  and  hydrogen  in  the  first 
five  members  of  the  methane  series  of  hydrocarbons  which  have  the 
general  formula  CnH2w+2.     (In  this  case  the  value  of  n  is  1,  2,  3,  4,  5.) 

14.  How  many  molecules  of  water  of  crystallization  were  present  in 
the    crystallized    salts    when    (a)    10    grams    of     cryst.   Na2CO3   lose 
6.29  grams  of  water  by  heating,  (6)  10  grams  of  cryst.  BaCl2  leave  after 
heating  a  residue  of  4.26  grams  anhydrous  salt,  (c)  crystallized  ZnSCX 
loses  after  heating  43.8  per  cent  of  its  weight? 

15.  Give  the  name  and  formula  of  a  compound  composed  of   (a) 
21.60  per  cent  sodium,  33.33  per  cent  chlorine,  45.07  per  cent  oxygen; 
(6)  70.13  per  cent  silver,  20.77  per  cent  oxygen,  9.10  per  cent  nitrogen; 
(c)  77.52  per  cent  lead,  4.49  per  cent  carbon,  17.88  per  cent  oxygen. 

16.  Ten  grams  of  the  chloride  of  a  univalent  element  x  contain  60.6 
per  cent  of  chlorine.     If  the  atomic  weight  of  chlorine  is  35.5  what  is  the 
atomic  weight  of  the  element  x,  and  what  is  its  name? 

17.  Find  the  anhydrides  of  the  following  substances  by  subtraction  of 
H20  (a)  HN02,  (b)  H2SO4,  (c)  HC1O3,  (d)  H3PO4,  (e)  Cu(OH)2,  (/)  A1(OH)3, 
(g)  LiOH,  («  H4SiO4,  (t)  Fe(OH)3,  (j)  HPO3.     Classify  these  anhydrides 
as  basic  and  acid  anhydrides. 

18.  1.251  grams  of  the  oxide  of  a  bivalent  element  when  reduced 
gave  1.00  gram  of  a  fine  metallic  powder.     What  is  the  atomic  weight  of 
the  metal,  and  what  is  its  name? 

19.  Calculate   the   simplest   formula   of    the   following    compounds: 
(a)  44.07  per  cent  mercury,  55.93  per  cent  iodine;  (b)  40  per  cent  calcium, 
12  per  cent  carbon,  48  per  cent  oxygen;  (c)   1.59  per  cent  hydrogen, 
22.22  per  cent  nitrogen,  76.16  per  cent  oxygen. 


CHAPTER  III 

EQUATIONS    INVOLVING    NO     OXIDATION     OR 
REDUCTION 

Equations  are  the  sentences  of  chemistry  as  they 
express  chemical  changes  which  take  place.  A  chemical 
change  or  chemical  reaction  is  a  molecular  phenomena 
in  which  the  composition  of  the  molecule  is  altered. 
Usually  a  chemical  reaction  is  brought  about  by 
mixing  solutions  of  two  different  substances.  The 
evidence  that ,  a  reaction  has  taken  place  is  (a)  the 
formation  of  a  precipitate  (insoluble  substance),  (6)  the 
formation  of  gas  bubbles  (gaseous  substance),  or  (c)  the 
change  or  formation  of  a  color  (differently  colored 
substance).  For  example,  (a)  by  mixing  a  solution  of 
potassium  iodide  and  lead  nitrate  a  yellow  precipitate 
is  produced,  for  the  following  reaction  has  taken  place: 

lead          nitrate     and     potassium     iodide 

will  give  \ 

potassium  nitrate 


and  lead    iodide 

(insoluble  and  yellow) 

If  (6)  hydrochloric  acid  (hydrogen  chloride)  is  added 
to  a  solution  of  sodium  sulfide  gas  bubbles  will  be 
observed,  for  the  following  exchange  has  taken  place: 

23 


24  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

hydrogen  chloride  and    sodium    sulfide 


will  give 


and 


hydrogen    sulfide 
(gaseous) 


If  (c)  a  few  drops  of  ferric  chloride  are  added  to  a 
solution  of  potassium  sulfocyanate  an  intensely  red  color 
will  be  produced,  due  to  the  formation  of  ferric 
sulfocyanate : 

ferric          chloride  and    potassium  sulfocyanate 


will  give 


and 


ferric  sulfocyanate 
(dark  red  colored) 


In  all  three  cases,  as  may  be  seen,  one  part  of  each 
molecule  has  changed  places  with  the  part  of  a  different 
molecule.  In  order  to  write  chemical  equations  for 
these  reactions  the  formulas  for  the  substances  which 
act  upon  each  other  are  placed  to  the  left  and  connected 
by  the  addition  sign  (  +  ),  while  the  reaction  products 
are  written  to  the  right  separated  by  the  sign  of  equality 
(  =  ),  which  means  that  the  number  and  kind  of  atoms 
on  both  sides  of  an  equation  must  be  equal.  Thus  an 
equation  demonstrates  the  principle  that  nothing  is  lost 
or  gained  in  the  course  of  transformation,  and  implies 
that  the  molecules  to  the  left  are  converted  into  those 
to  the  right.  In  reading  an  equation  the  plus  sign  is 
rendered  by  and,  and  the  equality  sign  by  give.  Balanc- 


EQUATIONS  25 

ing  of  an  equation  means  sim'ply  a  balancing  of  the 
number  of  atoms,  as  both  sides  must  be  equal. 

The  formulas  of  the  compounds  for  the  above  three 
reactions  are: 

(a)  Pb(N03)2  +  KI  =  KN03  +  PbI2 

(6)  HC1  +  Na2S  =  NaCl  +  H2S 

(c)  FeCl3  +  KCNS  =  KC1  +  Fe(CNS)3 

None  of  these  equations  are  balanced.  In  (a)  there 
are  two  (NO3)  radicals  to  the  left,  while  only  one  appears 
at  the  right,  and  only  one  I  atom  at  the  left,  to  two  I  on 
the  right.  To  balance  this  double  the  number  of  KI  on 
the  left,  and  KNO3  on  the  right,  and  the  correct  equation 
will  be 

(1)  Pb(NO3)2  +  2KI  =  2KNO3  +  PbI2. 

In  (6)  there  are  two  sodium  atoms  at  the  left  but 
only  one  at  the  right,  and  if  we  write  2NaCl  at  the 
right,  this  will  necessitate  2  chlorine  atoms  at  the  left, 
that  is  2HC1,  the  correct  equation  is  then : . 

(2)  2HC1  +  Na2S  =  2NaCl  +  H2S. 

In  (c)  there  are  three  chlorine  atoms  in  FeCl3  to  the 
left,  while  to  the  right  there  is  only  one  chlorine  atom 
in  KC1,  hence  writing  "3KC1"  to  the  right  balances 
the  chlorine  atoms  but  gives  three  potassium  atoms  to 
the  right,  and  as  only  one  potassium  atom  is  at  the 
left,  write  3KCNS  at  the  left  side  and  the  balanced 
equation  is : 

(3)  FeCl3  +  3KCNS  =  3KC1  +  Fe(CNS)3. 

These  are  now  balanced  molecular  or  non-ionic 
equations,  in  which  no  consideration  has  been  given 


26 


CHEMICAL   REACTIONS   AND    THEIR   EQUATIONS 


to  ionization.  The  number  and  kind  of  atoms  are 
identical  on  each  side  of  the  equality  sign.  In  balancing 
equations  the  correct  formula  for  each  compound 
involved  is  of  fundamental  importance,  for  the  first  step 
is  always  to  write  the  formulas  of  all  the  compounds, 
as  in  (a),  (&)  and  (c).  Knowledge  of  the  correct  formula 
naturally  requires  due  consideration  of  the  valency  of 
the  elements  and  radicals  concerned.  The  valency  of 
the  common  elements  must  be  memorized,  and  the 
valency  of  the  radicals  can  be  deduced  from  the  familiar 
formulas  of  a  few  acids  and  bases.  Thus  a  knowledge 
of  the  common  acids  and  bases  is  essential  for  the 
derivation  of  the  formulas  of  the  salts.  This  derivation 
is  demonstrated  in  the  following  equations: 


Base 

(4)  NaOH 

(5)  KOH 

(6)  Ca(OH)2 

(7)  Ba(OH)2 

(8)  Fe(OH)3 

(9)  A1(OH)3 

(10)  2NaOH 

(11)  3KOH 

(12)  Ca(OH)2 

(13)  3Ba(OH)2 

(14)  2Fe(OH)3 

(15)  Al(OH), 


All  these  equations  illustrate  that  a  base  and  an  acid 
will  give  a  salt  and  water.  This  type  of  reaction  is  called 
neutralization.  The  equations  above  are  given  in  the 
molecular  or  non-ionic  form,  in  the  ionic  form  they 
become : 


Acid 

=       Salt 

+  Water 

HC1 

=    NaCl 

+     H2O 

HN03 

=      KNO3 

+    H2O 

2HC1 

CaCl2 

+  2H2O  ' 

2HNO3 

=  Ba(NO3)2 

+  2H20 

3HC1 

=     FeCl3 

+  3H2O 

3HNO3 

=    A1(N03)3 

+  3H2O 

H2SO4 

Na2SO4 

+  2H2O 

H3PO4 

=  K3PO4 

+  3H2O 

H2S04 

=  CaSO4 

+  2H20 

2H3PO4 

=  Ba3(PO4)2 

+  6H2O 

3H2SO4 

=  Fe2(S04)3 

+  6H20 

H3PO4 

=  A1PO4 

+  3H20 

EQUATIONS  27 

(4a)  Na+  +  OH-  +  H+  +  Cl~  =  Na+  +  Cl-  +  H2O 
(5a)  K+  +  OH-  +  H+  +  NO3-  =  K+  +  NO3-  +  H2O 

(lOa)  2Na+  +  2OH~  +  2H+  +  SO4~  =  2Na+  + 

SO4-  +  2H2O 

(lla)  3K+  +  3OH-  +  3H+  +  PO4-       =    3K+    + 

PO4—     +  3H2O  and  so  on. 

An  examination  of  the  ionic  equations  (4a),  (5a),  (lOa), 
(lla)  shows  that  some  ions  have  not  changed.  E.g.  in 
(4a)  the  Na+  and  Cl~  on  the  left  appear  also  on  the  right, 
and  therefore  have  undergone  no  change.  Similarly 
the  K+  and  NO3~  of  equation  (5a)  remain  unchanged. 
The  purpose  of  an  equation  is  to  show  a  chemical  reac- 
tion, that  is  any  chemical  change,  and  if  all  the  ions  which 
have  not  changed  are  cancelled,  there  remains  from  the 
above  equations 

(16)  OH-  +  H+  =  H20 

This  is  the  general  equation  for  neutralization,  and 
defines  neutralization  as  the  combination  of  one  hydroxyl 
ion  with  one  hydrogen  ion  to  give  one  molecule  of  water. 
By  using  this  universal  statement  all  the  particular 
cases  can  readily  be  found  by  adding  to  the  negative 
hydroxyl  ion  (OH~)  the  positive  metal  of  any  base,  and 
to  the  positive  hydrogen  ion  (H+)  the  negative  non- 
metal  or  radical  of  any  acid.  From  the  proportion  1  :  1 
it  is  evident  that  any  monoacid  base  requires  one,  a  di- 
acid  base  =  two,  a  triacid  base  =  three  hydrogen  ions; 
while  a  monobasic  acid  requires  one,  a  dibasic  acid  =  two, 
a  tribasic  acid  =  three  hydroxyl  ions.  Thus  in  equa- 
tion (10)  two  molecules  of  a  monoacid  base  are  needed 
for  neutralization  of  one  molecule  of  a  dibasic  acid. 

If  only  one  of  the  reaction  products  is  known  in  a 
reaction  the  other  reaction  product  is  found  by  sub- 


28  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

tracting  the  atoms  of  the  right  from  those  on  the  left. 
Thus,  if  the  unfinished  equation  is  BaCl2  +  Na2SO4  = 
BaSO4  +  ?,  the  following  scheme  will  give  the  unknown 
reaction  product: 

On  the  left        Ba  C12  Na2  S  O4 
On  the  right     Ba   -  S  Q4 

Remainder        —  C12  Na2   • 

These  remaining  two  chlorine  and  two  sodium  atoms 
naturally  give  two  molecules  of  sodium  chloride,  hence 
the  completed  equation  is 

(17)  BaCl2  +  Na2SO4  =  BaSO4  +  2NaCl 
Writing  this  equation  in  the  ionic  form: 

Ba++  +  2C1-  +  2Na+  +  SO4~  =  BaSO4  +  2Na+  + 

2C1- 

and  cancelling  those  ions  which  remain  unchanged  (Gl- 
and Na+)  the  equation  becomes: 

(18)  Ba++  +  SO4—  =  BaS04 

This  is  again  a  general  statement,  namely  that  every 
soluble  barium  salt  (furnishing  the  barium  ion)  with  any 
soluble  sulfate  (which  furnished  the  sulfate  ion)  will  give 
a  precipitate  of  barium  sulfate  (which  is  insoluble).  It 
is  evident  that  an  ionic  equation  covers  many  cases, 
for  one  may  take  as  the  soluble  barium  salt  either  BaCl2, 
Ba(NO3)2,  etc.,  and  for  the  soluble  sulfate  either  Na2SO4, 
K2SO4,  etc.,  but  it  refers  only  to  one  specific  reaction, 
e.g.,  the  formation  of  barium  sulfate.  A  non-ionic 
or  molecular  equation  covers  only  one  particular  case, 
for  from  such  an  equation  as  (17)  we  can  not  make  the 
deductions  which  were  possible  from  (18). 

Balancing  Molecular  Equations. — An  unfinished  equa- 
tion may  be  given  in  which  there  are  more  atoms  in  the 


EQUATIONS  29 

reaction  product  than  in  the  reacting  substances.  A 
simple  case  is  CuO  +  KC1  =  CuCl2  +  ?,  and  as  2C1 
can  not  be  subtracted  from  1  Cl,  the  number  of  HC1 
must  be  doubled: 

On  the  left        Cu  O  2H  2C1 
On  the  right     Cu  C12 


Remainder        -    O  2H     - 
which  will  form  one  molecule  of  water,  thus: 
(19)  CuO  +  2HC1  =  CuCl2  +  H2O 

The  problem  becomes  more  difficult  when  two  or  more 
compounds  are  formed,  and  in  this  case  greater  knowledge 
of  chemistry  is  required.  For  example  an  experiment 
shows  that,  on  adding  a  solution  of  sodium  carbonate  to  a 
solution  of  aluminum  sulfate,  a  white  precipitate  as  well 
as  gas  bubbles  are  formed.  Hence  the  reaction  products 
are  (a)  an  insoluble  compound,  and  (6)  a  gaseous  com- 
pound. As  neither  the  Al,  the  Na,  nor  the  SO4  radical 
form  gaseous  compounds,  but  as  the  carbonates  yield 
readily  CO2,  the  unknown  gas  must  be  CO2  and  the  un- 
finished equation  is 

A12(SO4)3  +  Na2CO3  =  CO2  +  ? 

The  white  precipitate,  which  is  the  insoluble  compound 
formed  in  this  reaction,  must  be  a  compound  of  aluminum 
for  all  sodium  compounds  are  soluble.  At  first  it  appears 
to  be  the  carbonate,  but  as  carbonic  acid  is  given  off,  this 
can  not  be  the  case.  The  other  possible  compounds  to 
consider  would  be  the  oxide  and  the  hydroxide  of  alumi- 
num, for  in  this  reaction  the  other  insoluble  aluminum 
compounds  like  the  phosphates,  silicates,  etc.  can  not 
be  regarded  as  possible  reaction  products,  as  neither  of 
these  acid  radicals  are  present.  The  reaction  occurred  in 


30  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

watery  solution  hence  the  hydroxide  is  probably  formed; 
if  so,  then  water  must  have  taken  part  in  the  reaction, 
and  the  unfinished  equation  becomes: 

A12(S04)3  +  Na2C03  +  H20  =  CO2  +  A1(OH)3  +  ? 
The  next  metal  to  account  for  in  this  equation  is  sodium, 
which  loses  the  carbonate  radical,  and  must  combine 
with  the  sulfate  radical  to  form  sodium  sulfate: 

A12(S04)8  +  Na2C03  +  H20   =   CO2   +  A1(OH)3  + 

Na2SO4 

An  examination  of  this  unbalanced  equation  shows  that 
there  are  two  aluminum  to  the  left  and  hence  2A1(OH)3 
must  be  formed.  There  are  3SO4 — radicals  to  the  left, 
thus  3Na2SO4  are  produced — this  in  turn  requires 
3Na2CO3,  and  these  three  molecules  will  yield  3CO2. 
Finally  as  2A1  (OH)3  are  formed,  there  are  2  X  3  =  6 
hydrogen  atoms  necessary  and  these  are  present  in 
3H2O.  Making  these  corrections,  the  finished  and 
balanced  equation  is: 

(20)  A12(SO4)3  +  3Na2CO3  +  3H2O   = 

3CO2  +2A1-  (OH),  +  3Na2SO4 

As  a  test  of  correctness  the  following  checking  scheme  is 
used: 

Al       (SO4)       Na       C         O        H 
To  the  left  2363       9+3       6 

To  the  right  2  3  6         36+66 

From  this  we  see  that  one  atom  of  oxygen  from  each  of 
the  three  carbonate  radicals  reappears  in  a  molecule  of 
the  hydroxide. 

To  write  equation  (20)  in  the  ionic  form  the  soluble 
salts  are  written  as  ions,  while  those  ions  which  appear 
on  both  sides  are  cancelled,  and  there  remains: 

(21)  2A1+++  +  3CO3—  +  3H2O  =  2A1(OH)3  +  3CO2 


EQUATIONS  31 

The  general  rule  drawn  from  this  equation  is  that  every 
soluble  aluminum  salt  will  react  with  soluble  carbonates 
and  water  to  give  insoluble  aluminum  hydroxide  and 
gaseous  carbon  dioxide.  It  is  again  evident  that 
equation  (20)  referred  to  a  particular  case,  while  equation 

(21)  refers  to  a  specific  reaction  which  covers  a  number 
of  particular  cases,  for  the  aluminum  sulfate  may  be 
substituted  by  aluminum  chloride,  or  aluminum  nitrate, 
etc.,  and  the  sodium  carbonate  by  potassium  carbonate, 
or  lithium  carbonate,  etc. 

Balancing  Ionic  Equations. — In  balancing  an  unfin- 
ished ionic  equation  the  charges  on  the  ions  must  also  be 
balanced.  As  a  rule  the  balancing  is  simple  if  the 
charges  are  balanced  first,  before  the  atoms  are  balanced. 
Thus  in  Fe+++  +  CO3 —  =  ?;  there  are  three  positive 
charges  on  the  iron,  and  two  negative  charges  on  the 
carbonate  radical,  and  to  make  their  sum  zero  converse 
and  add  2(+3)  and  3 (—2),  that  is  two  ferric  ions 
(together  six  positive  charges)  and  three  carbonate  ions 
(together  six  negative  charges),  therefore: 

(22)  2Fe+++  +  3CO3—  =  Fe2(CO3)3 

It  is  not  necessary  that  the  sum  of  the  charges  on  each  side 
of  an  equation  be  zero,  as  in  (22),  it  can  be  any  integer, 
provided  this  sum  is  the  same  on  both  sides  of  the  equation. 

Thus  in  (23)    Pb++  +  H2S  =  PbS  +  2H+ 

there  are  two  positive  charges  on  the  left  which  equal 
the  two  positive  charges  on  the  right.  Similarly  in 
the  equation 

(24)  Zn(OH)2  +  2OH-  =  ZnO2—  +  2H2O 

there  is  on  both  sides  of  the  equations  an  excess  of  two 
negative  charges,  while  in 

(25)  Cr2O7—  +  2OH-  =  2CrO4—  +  2H2O 


32  CHEMICAL   REACTIONS    AND   THEIR    EQUATIONS 

the  sum  is  four  negative  charges  on  both  sides.  The  sum 
of  the  ionic  charges  may  thus  be  different  for  different 
reactions,  .but  must  be  equal  on  both  sides  of  the 
equation. 

The  general  meaning  of  the  last  four  ionic  equations 
is  that: 

(22)  any  soluble  ferric  salt  with  any  soluble  carbonate 
will  form  an  insoluble  precipitate  of  ferric  carbonate; 

(23)  any  soluble  lead  salt  and  hydrogen  sulfide  will  give 
insoluble  lead  sulfide  and  liberate  hydrogen  ions  ; 

(24)  zinc    hydroxide    dissolves    in   bases    (OH—)    and 
forms  zincates  and  water; 

(25)  soluble  bichromates  with  bases  give  soluble  chromates. 
Rules  for  Finishing  and  Balancing  Equations. — The 

rules  in  balancing  and  finishing  equations  are  summarized 
as  follows: 

(A)  Unfinished  equations  in  which  only  the  reacting 
substances  are  given,  and  none  or  one  of  the  reaction 
products  (compare  equations  (17),  (18),  (19),  (20)).     In 
such  cases  there  is  usually  an  exchange  of  the  positive 
atoms  or  radicals,  according  to  the  type  MN  +  M'N'  = 
M'N  +  MN',  where  M  and  M'  are  positive  elements,  and 
N  and  N'  negative  elements  or  radicals.     The  finishing 
of  such  problems  requires: 

First,  the  correct  formulas  for  the  reacting  substances, 
(to  the  left)  and  the  correct  formulas  of  the  reaction 
products  (to  the  right).  The  reaction  products  are 
usually  gotten  by  exchanging  the  positive  or  the  negative 
constituents  of  the  molecules,  giving  due  consideration 
to  the  valency. 

Second,  balance  the  atoms,  as  given  under  B. 

(B)  Balancing  equations  means  equalizing  the  number 
and  kind  of  atoms  on  both  sides  of  the  equation:  Begin 
with  any  symbol  on  the  left  and  check  the  same  number 


EQUATIONS  33 

of  symbols  on  the  right — if  the  number  is  unequal,  select 
the  highest  number  on  either  side  of  the  equation,  and  write 
the  same  number  before  the  formula  containing  the 
symbol  on  the  opposite  side  of  the  equation.  Hydrogen 
and  oxygen  atoms  are  often  balanced  by  writing  H2O  on 
the  opposite  side  of  the  equation. 

(C)  Balancing  ionic  equations  requires: 

First,  equalizing  the  charges  on  both  sides  of  the  equa- 
tion: the  sum  of  the  charges  on  the  left  must  equal  the 
sum  of  the  charges  on  the  right.  If  the  sums  are  unequal 
and  they  cannot  be  made  equal  in  the  way  shown  for 
equation  (22),  the  addition  of  H+  or  OH~  is  necessary, 
depending  upon  whether  the  reaction  occurs  in  acid  or 
alkaline  solution. 

Second,  balance  the  atoms,  as  given  under  B. 

An  ionic  equation  is  always  a  general  statement  of  the 
chemical  properties  of  an  ion,  while  the  non-ionic  or 
molecular  equation  is  a  particular  statement  of  a  definite 
case,  or  of  the  weight  relations  in  a  certain  reaction,  for 
by  introducing  the  respective  mass  of  the  atoms  and 
molecules,  the  proportions  of  weight  of  the  reacting 
substances  are  found.  The  following  questions  involve 
some  of  these  weight  relations  and  with  the  aid  of  the 
tables  in  the  appendix  should  offer  no  difficulty. 

QUESTIONS 

1.  Define  (a)  equation,  (b)  ionic  equation,  (c)  neutralizatio 

2.  What  is  the  cost  of  materials  for  preparing  100  kg.  of        p<         ut 
solution  of  HC1  from  NaCl  and  H2SO4,  if  the  kg.  of  NaCl  c<,^s     p,  and 
the  kg  H2SO4,  6^? 

3.  How  many  grams  of  KBr  and  AgNO3  are  needed  to  make  half  a 
kg.  of  AgBr? 

4.  Eighty  grams  of  a  solution  of  sulphuric  acid  are  sufficient  to  dissolve 
22  grams  of  cupric  oxide.     What  was  the  percentage  by  weight  of  sul- 
phuric acid  in  the  original  solution? 

5.  Three  hundred  and  ten  grams  of  borax  crystals  are  dissolved  in  a 

3 


34  CHEMICAL   REACTIONS   AND    THEIR    EQUATIONS 

little  water  and  concentrated  sulphuric  acid  is  added.  How  many  grams 
of  H2SO4  should  be  used,  and  how  many  grains  of  boric  acid  are  obtained  ? 
Na2B407.10H20  +  H2SO4  =  Na2SO4  +  4H3BO3.  +  5H2O. 

6.  How  many  kilograms  of  KNO3  can  be  made  from  a  metric  ton  of 
NaNOs,  and  how  many  kilograms  of  KC1  must  be  used?     (A  metric  ton 
equals  1,000  kg.) 

7.  Calculate  the  loss  in  weight  if  15  grams  of  marble  are  heated  until 
the  change  is  complete. 

8.  To  32  grams  of  a  solution  of  sulphuric  acid  an  excess  of  BaCl2  was 
added.     The    BaSO4   obtained    weighed    11.43   grams.     Calculate  the 
percentage  of  sulphuric  acid  in  the  sulphuric  acid  solution. 

9.  How  many  grams  of  HF1  can  be  made  from  20  grams  of  fluorspar 
(CaFl2),  and  how  many  grams  of  sulphuric  acid  must  be  added?     What 
will  be  the  volume  of  HF1  gas,  if  one  mole  occupies  22.4  liters? 

10.  How  many  grams  of  NH3  can  be  obtained  from  50  grams  of 
ammonium  chloride,  and  how  many  grams  of  calcium  hydroxide  are 
necessary?     What  will  be  the  volume  of  the  NH3  gas? 

11.  How  many  grams  of  nitric  acid  will  be  obtained  by:     (a)  heating 
250  kg.  of  sodium  nitrate  with  sulphuric  acid;  (6)  heating  250  kg.  of 
potassium  nitrate  with  sulphuric  acid? 

12.  How  many  grams  of  crystallized  cupric  nitrate  (containing  3  mol. 
H2O)  must  be  heated  to  redness  to  make  50  grams  of  cupric  oxide? 

13.  How  many  grams  of  nitric  acid  are  needed  to  convert:  (a)  500 
grams  sodium  hydroxide  into  sodium  nitrate;  (6)  500  grams  potassium 
hydroxide  into  potassium  nitrate;  (c)  500  grams  calcium  hydroxide  into 
calcium  nitrate;  (d)  500  grams  calcium  oxide  into  calcium  nitrate;  (e)  500 
grams  barium  hydroxide  into  barium  nitrate? 

14.  A  solution  contains  exactly  40.00  grams  of  sodium  hydroxide  in 
1   liter.      Calculate  the  quantities  by  weight  of   (a)  HC1  (b)   HNOs, 
(c)  H2SO4,  (d)  H5PO4  which  will  be  required  to  neutralize  10  c.c.  of  this 
solution. 

15.  In  ascertaining  the  strength  of  a  dilute  solution  of  HC1,  50  c.c.  were 
measured  out  and   neutralized  with   a  solution  of  sodium  hydroxide, 
containing  0.003  grams  NaOH  in  1  c.c.     Thirty-five  cubic  centimeters 
of  this  solution  was  required.     What  is  the  strength  of  the  HC1?     How 
many  grams  HC1  are  in  1  c.c.? 

16.  Forty    cubic    centimeters    of  a  solution  of  potassium  hyroxide 
containing  0.01  grams  of  KOH  in  1  c.c.  were  required  to  neutralize 
40  c.c.  of  a  solution  of  H2SO4.     How  many  grams  of  sulphuric  acid  did 
15  c.c.  of  the  sulphuric  acid  solution  contain? 

17.  Three  hundred  and  eighty  grams  of  mercuric  chloride  are  dissolved 
in  water.     How  many  grams  of  Kl  must  be  added  to  precipitate  all  the 
mercury?     How  many  grams  of  HgI2  will  be  obtained? 

18.  If  5  grams  of  crystallized  copper  sulf ate,  CuSO4.5H2O,  are  dissolved 


EQUATIONS  35 

in  water,  how  many  grams  of  H2S  are  necessary  to  precipitate  all  the 
copper  as  copper  sulfide?  How  many  grams  of  FeS  and  25  per  cent 
HC1  are  required  to  generate  enough  H2S  for  this  purpose? 

19.  How  would  3^011  make  a  normal  solution  of  the  following  sub- 
stances: (a)  HC1,  (h)  HNO3,  (c)  H2SO4,  (d)  NaOH,  (e)  NH4OH?     How 
would  you  make  a  N/10  solution  of  (/)  Ca(OH)2,  (g)  NaCl,  (h)  SO2; 
a  2N  solution  of  (i)  HF1,  (j)  H3PO4,  (it)  FeSO4? 

20.  Answer  the  following  problems  by  inspection : 

(a)  How  much  KOH  will  be  required  to  neutralize  1  liter  N  solution  of 
HNO3,  1  liter  N  solution  of  H2SO4,  1  liter  N  solution  of  H3PO4?  (6) 
How  many  grams  of  Ca(OH)2? 

21.  Write  the  equations  for  preparation  of  the  following: 
(a)  Cupric  oxide  from  cupric  carbonate; 

(6)  Cupric  chloride  from  cupric  oxide ; 

(c)  Ferric  hydroxide  from  ferric  sulphate; 

(d)  Ferrous  sulfide  from  ferrous  sulphate; 

(c)  Aluminum  carbonate  from  aluminum  nitrate; 
(/)  Aluminum  hydroxide  from  aluminum  chloride. 
State  in  each  case  which  substances  are  necessary  for  the  reaction  and 
which  substances  are  precipitated. 

22.  How  would  you  make  the  following  compounds: 
(a)  Cupric  chloride  from  cupric  sulfate? 

(6)  Cupric  sulphate  from  cupric  chloride? 

(c)  Ferrous  sulphate  from  ferrous  sulfide? 

(d)  Ferric  chloride  from  ferric  sulphate? 

(e)  Aluminum  sulfate  from  aluminum  chloride? 
(/)  Aluminum  chloride  from  aluminum  sulfate? 

As  none  of  these  compounds  are  insoluble,  in  which  way  should  they 
be  prepared?  Base  your  answers  on  equations  which  show  possible 
reactions. 

23.  You  are  given  a  quantity  of  lead  acetate  solution.     How  would 
you  prepare:  (a)  lead  chloride,  (6)  lead  nitrate,  (c)  lead  sulfate,  (d)  lead 
chromate,  (e,  lead  oxide,  (f)  lead  carbonate  from  this  solution? 

24.  Starting  with  lime  CaO  how  would  you  make   (a)   CaCl2,    (fe) 
CaSO4,  (c)  Ca(NO3)2,  (d)  CaCO3,  (e)  solid  Ca(OH)2? 

25.  Describe  at  least  three  ways  of  making  ferric  chloride  from  ferric 
sulphate. 

26.  Write   the   equations  for   the  formation   of:  (a)    Prussian   blue 
Fe4[Fe(CN)6]3  from  ferric  ions  and  ferrocyanide  ions;  and  (6)  Thurnbull's 
blue  Fe3[Fe(CN)8]2  from  ferrous  ions  and  ferricyanide  ions. 

27.  Finish  the  following  equations: 
(a)  Zn++  -i-  H2S  = 

(6)  NiCO3  +  H2SO4  = 
(c)  Ni++  -f  (NH4)2S  = 


36  CHEMICAL   REACTIONS   AND   THEIR   EQUATIONS 

(d)  HgN03  +  Cl-  = 

(e)  A1+++  +  OH'  = 

(/)  A1(OH)8  =  A1203  +  ? 
(g)  FeS  -f  H+  =  H2S 
(h)  CaF2  +  ?  =  CaS04  +  ? 
(i)  Ba++  +  ?  =  BaC03  +  ? 
0')  A1+++  +  ?  =  A1(OH)8  +  H+ 
(fc)  Ca(OH)2  +  ?  =  Ca(SH)2  +  H2O 
(0  NaCl  +  NH4HC03  =  NaHCO3  +  ? 
(m)  CuO  +  ?  +  H20  =  Ba(OH)2  -f  CuS 
(n)  CoCl2.6H2O  =  Co++  +  ?  +  ? 
(o)  CoCl2.6H2O  =  CoCl2  -f  ? 
(p)  MnCl2  +  (NH4)2S  =  Mn(SH)2  +  ? 
(q)  6KCN  +  FeS04  =  K4Fe(CN)6  +  ? 
28.  Balance  the  following  equations : 
(a)  CaC03  +  HN03  =  Ca(NO3)2  +  H2O  +  CO2 
(6)  SiCl4  +  NH3  =  Si(NH2)4  +  NH4C1 

(c)  SiS2  +  H20  =  2H2S  +  Si02 

(d)  KaSiF.  +  KOH  =  KF  +  K4SiO4 

(e)  SiF4  +  H20  =  H4Si04  +  H2SiF6 
(/)  PC16  +  AsF3  =  PF5  +  AsCl3 


CHAPTER  IV 

EQUATIONS  INVOLVING  OXIDATION  AND 
REDUCTION 

Oxidation  and  Reduction. — Oxidation  and  reduction 
are  defined  in  a  previous  chapter  as  the  increase  or  de- 
crease in  the  valence  numbers.  In  an  equation  this 
increase  or  decrease  of  the  valence  number  must  also  be 
balanced.  Thus  an  increase  of  two  in  the  valence 
number  of  one  element  must  be  balanced  by  a  decrease 
of  two  in  the  valence  number  of  some  other  element. 
In  other  words,  oxidation  cannot  take  place  unless  there 
is  a  reduction  in  some  other  element.  In  its  simplest 
form  this  principle  is  illustrated  in  the  displacement 
reaction : 

(26)  Cu++  +  Fe  =  Cu  +  Fe++ 

In  this  case  the  metallic  iron  (Fe  =  valence  number  O) 
acquired  two  positive  charges  from  the  cupric  ion  (Cu++ 
=  valence  number  +2)  and  was  oxidized  (increased  its 
valence  number)  to  ferrous  ion  (Fe++  =  valence  number 
+2),  while  at  the  same  time  the  cupric  ion  was  reduced 
(decreased  its  valence  number)  to  metallic  copper  (Cu  = 
valence  number  O).  The  experiment  is  readily  per- 
formed by  placing  a  piece  of  blank  iron  in  a  solution  of 
any  copper  salt,  e.g.,  blue  vitriol,  and  the  iron  will  be 
covered  by  a  film  of  finely  divided  copper  in  the  form  of 
a  brownish  precipitate.  If  sufficient  time  is  given  for 
the  reaction  to  take  place  all  of  the  copper  may  be  pre- 
cipitated, and  the  blue  color  of  the  solution  will  change 
to  the  greenish  color  of  the  ferrous  ion.  In  this  case 

37 


35  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

the  metallic  iron  has  replaced  the  copper  (see  also 
Chapter  5). 

Not  only  with  positively  charged  elements,  but  also 
with  negatively  charged  elements  such  a  displacement 
takes  place: 

(27)  C12  +  2Br-  =  2C1~  +  Br2 

In  this  case  chlorine  (O)  was  reduced  to  chloride  ion  ( —  1), 
and  the  bromide  ion  (  —  1)  was  oxidized  to  bromine  (O). 
It  should  be  noted  that  in  equation  (26)  the  transforma- 
tion of  an  element  to  its  ion  was  oxidaton,  while  in 
equation  (27)  this  transformation  was  a  reduction,  in  the 
first  case  the  valence  number  increased,  in  the  second 
case  the  valence  number  decreased. 

Metals  and  Non-metals. — As  a  general  rule  metals 
are  oxidized,  and  non-metals  are  reduced  to  binary1 
compounds,  and  oxidized  to  tertiary2  compounds.  An 
exception  to  this  rule  is  the  direct  oxidation  of  non- 
metals  by  oxygen,  e.g.,  burning  sulfur,  phosphorus,  or 
carbon  in  air,  as  in  these  reactions  binary  compounds 
are  formed,  e.g.  S02,  P205,  CO,  or  CO2.  The  reason  for 
this  is  found  in  the  assumption  that  oxygen  is  always 
considered  negative,  or  —2,  and  therefore,  in  non-metal 
oxides,  the  non-metal  is  positive  with  regard  to  oxygen; 
while  in  non-metal  hydrides  (e.g.,  H2S,  NH3,  CH4)  and 
the  binary  salts  (e.g.,  KC1,  CaS,  A1N,  FeC2)  the  non-metal 
is  always  negative,  as  hydrogen  is  positive  (+1)  and  all 
metals  are  positive. 

In  the  equation 

(28)  2HgO  =  2Hg  +  02 

the  mercury  of  mercuric  oxide  was  reduced,  the  valence 

1  Binary  compounds  have  only  two  kinds  of  atoms,  e.g.,  NaCl,  FeCl2. 

2  Tertiary  compounds  have  three  different  kinds  atoms,  e.g.,  KCN, 
BaSO4,  Na3PO4. 


EQUATIONS  39 

number  being  decreased  from  2  to  0,  while  the  oxygen  of 
mercuric  oxide  was  oxidized,  the  valence  number  being 
increased  from  —2  to  0.  However,  in  the  equation 

(29)  4Na  +  02  =  2Na2O 

metallic  sodium  (O)  is  oxidized  to  sodium  oxide  (Na  = 
+  1),  while  the  oxygen  gas  (02)  was  reduced  to  an  oxide 
(-2). 

Balancing  Oxidation  and  Reduction. — As  seen  in  these 
equations,  oxidation  is  exactly  counter-balanced  by  re- 
duction, for  the  increase  in  valence  number  is  the  re- 
ciprocal of  the  decrease  in  valence  number,  in  other 
words,  if  the  increase  is  added  to  the  decrease  the  sum  is 
zero.  Thus  in  equation: 

(26)  oxidation  of  iron  to  ferric  ion  =  increase  of  two, 
reduction  of  cupric  to  copper  ion  =  decrease  of  two; 

(27)  oxidation  of  two  bromide  ions  to  bromine  =  increase 

of  2  X  1  =  2 

reduction  of  chlorine  to  chloride  ions  =  decrease  of 
2  X  -1  =  -2 

(28)  oxidation  of  two  oxides  to  oxygen  =  increase  of 

2X2  =  4 
reduction  of  two  mercuric  to  mercury  =  decrease 

of  2  X  2  =  -4 

(29)  oxydation  of  four  sodium  =  increase  of  4  X  1  =4 
reduction  of  two   oxygen  to  oxide  =  decrease  of 

2  X  2  =  -4 

Sum  Is  Zero. — The  sum  of  this  increase  and  decrease 
is  always  zero  in  a  correct  equation.  A  more  complex 
example  is: 

(30)  Sn  +  2HN03  +  H2O  =  H4Sn04  +  N2O3. 


40  CHEMICAL   REACTIONS    AND    THEIR    EQUATIONS 

In  this  equation  metallic  tin  (O)  has  been  oxidized  to  a 
stannic  compound  (4)— stannic  acid,  while  the  nitrogen 
of  nitric  acid  (5)  has  been  reduced  to  nitrous  oxide  (3), 
thus  the  tin  has  increased  by  4,  and  two  nitrogen  atoms 
have  each  decreased  by  2,  and  2  (—2)  =  —4. 

This  increase  or  decrease  of  the  valance  number  is 
perhaps  best  illustrated  by  writing  the  valence  numbers 
which  change  under  the  respective  symbols: 

(31)  2HBr  +  H2S04  =  SO2  +  Br2  +  2H2O 

(-1)  (6)  (4)        (0) 

In  this  reaction  two  bromine  atoms  (  —  1)  of  hydrobromic 
acid  have  been  oxidized  to  free  bromine  (0),  and  the 
sulfur  (6)  of  sulfuric  acid  has  been  reduced  to  sulfur  (4) 
of  sulfur  dioxide.  However  in 

(32)  2HBr  +  MnO2  +  H2SO4  =  MnSO4  +  Br2  +  2H20 

(-1)         (4)  (2)  (0) 

not  the  sulfur  of  sulfuric  acid,  but  the  manganese  (4)  of 
manganese  dioxide  has  been  reduced  to  manganese  (2) 
of  manganous  sulfate.  Writing  the  later  equation  in 
the  ionic  form : 

(33)  2Br~  +  MnO2  +  4H+  =  Mn++  +  Br2  +  2H2O 

illustrates  the  rule  that  any  soluble  bromide  in  the 
presence  of  hydrogen  ion  (any  acid)  will  give,  with 
manganese  dioxide,  a  manganous  compound  and  free 
bromine. 

In  balancing  an  unfinished  equation  it  is  always 
advantageous,  and  in  many  cases  absolutly  necessary, 
to  strictly  follow  the  rules. 

The  Three  Rules. — First,  balance  the  valence  numbers 
of  the  elements.  Write  under  each  symbol  its  valence 
number,  then  observe  which  elements  are  oxidized  or 


EQUATIONS  41 

reduced.  The  oxidation  must  be  the  reciprocal  of  the 
reduction,  for  if  the  element  A  has  been  oxidized  from 
2  to  4,  (increased  two  steps)  and  the  element  B  has  been 
reduced  from  3  to  0,  (decreased  three  steps)  then  3  atoms 
of  A  will  counterbalance  2  atoms  of  B,  for  3X2  =  6, 
and  2  X  (-3)  =  -6,  and  6  +  (-6)  =  0. 

Second,  balance  the  ionic  charges  on  both  sides. 
Write  on  each  side  the  sum  of  the  ionic  charges  of  all 
the  ions.  This  sum  may  be  either  +  ,  —  ,  or  0,  but  must 
be  the  same  on  each  side.  If  the  sum  is  not  equal  and 
the  reaction  occurs  in  an  acid  solution,  add  the  required 
number  of  hydrogen  ions  (H+)  to  the  proper  side,  and, 
if  it  occurs  in  alkaline  solution,  add  hydroxyl  ions 
(OH~),  until  the  sum  becomes  equal. 

Third,  balance  the  atoms.  Their  kind  and  number 
must  be  equal  on  both  sides. 

Strict  adherence  to  the  sequence  of  these  three  steps  is 
essential  for  a  speedy  solution  of  the  most  complex 
problems;  for  any  attempt  to  balance  the  number  of 
atoms  before  the  valence  numbers  and  ionic  charges  are 
balanced  may  result  in  failure  or  an  incorrect  solution. 
A  few  examples  will  illustrate  the  method. 

Examples. — Problem  1. — Write  an  equation  for  the 
oxidation  of  a  ferrous  salt  to  a  ferric  salt  by  potassium 
permanganate  in  acid  solution.  Supposing  the  per- 
manganate is  reduced  to  a  manganous  salt,  we  write  the 
four  symbols  for  the  ions  and  place  under  each  symbol 
the  respective  valence  numbers: 

MnO4-  +  Fe++  =  Mn++  +  Fe+++ 
7  22  3 


Connecting  the  corresponding  valence  numbers  by  the 


42  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

dotted  lines  shows  that  manganese  has  been  reduced 
from  7  to  2  (5  steps),  while  the  iron  is  oxidized  from  2 
to  3  (1  step).  In  order  that  oxidation  and  reduction  are 
balanced,  the  factors  are  reversed :  1  X  5  steps  =  5X1 
step,  which  means  that  1  molecule  of  permanganate  is 
sufficient  to  oxidize  5  molecules  of  a  ferrous  compound 
and  the  unfinished  equation  becomes: 

MnO4~  +  5Fe++  =  Mn++  +  5Fe+++ 

The  proportion  of  IMn  to  5Fe  must  be  kept  throughout 
the  succeeding  operations — it  may  be  doubled  (2:10)  or 
tripled  (3:10) — but  the  proportion  of  1  to  5  must  be 
preserved. 

The  second  step  is  balancing  the  ionic  charges.     They 
are  counted  and  added  together  separately  for  each  side 
of  the  equation,  thus: 
on  the  left : 

one  negative  charge  on  the  permangante  =  —    1 
5X2  positive  charges  on  the  iron  .  10 

therefore  excess  of  positive  charges  9 

on  the  right: 

two  positive  charges  on  the  manganese      =      2 
5X3  positive  charges  on  the  iron  =     15 

a  total  of  positive  charges  17 

As  the  number  of  charges  on  both  sides  must  be  equal, 
for  nothing  can  be  gained  or  lost  during  a  chemical 
reaction,  some  other  substance  must  have  reacted  which 
furnished  either  8  positive  charges  to  the  left,  or  yielded 
8  negative  charges  to  the  right  side  of  the  equation. 
The  positive  charges  are  usually  furnished  by  H+ 
(acids),  and  the  negative  charges  by  OH~  (bases),  and 


EQUATIONS  43 

as  it  was  specified  that  the  reaction  occurred  in  acid 
solution,  8H+  must  be  added  to  the  left  in  order  to 
balance  the  charges: 


MnO4-  +  5Fe++  +  8H+  =  Mn++  +  5Fe+++. 

The  sum  of  the  ionic  charges  is  thus  equal  on  both  sides 
for  (-1)  +  5(2)  +  8(1)  =  (2)  +  5(3)  =  17;  and  the 
first  and  second  step  of  balancing  is  accomplished.  It 
remains  to  balance  the  atoms.  Checking  off  the  atoms 
leaves  4  oxygen  atoms  and  8  hydrogen  atoms  on  the 
left  which  will  naturally  give  4  molecules  of  water. 
Therefore  the  completed  equation  is 
(34)  MnO4~  +  5Fe++  +  8H+  =  Mn++  +  5Fe+++  +  4H2O. 

This  ionic  equation  is  the  general  statement  of  the 
problem  and  enables  the  construction  of  many  equations 
for  particular  cases  by  simply  supplying  the  ions  with 
the  corresponding  radicals  or  elements.  In  other  words, 
the  permanganate  (Mn04~)  can  be  any  soluble  per- 
manganate, either  of  sodium,  potassium,  etc.;  the  ferrous 
salt  can  be  any  soluble  salt,  either  sulfate,  chloride, 
nitrate,  etc.;  and  the  acid  can  be  either  HC1,  H2SO4, 
HNO3,  etc. 

Problem  2.  —  A  nitrite  solution  is  oxidized  to  a  nitrate 
solution  by  an  acidified  bichromate  solution  : 

First,  the  valence  numbers  are  written  under  the  ions 


NO2-  - 

+-  Cr,O7- 

=  N03-  - 

f  2Cr+++ 

3 

2(6) 

5 

2(3) 

=  12 

j 

=  6 

The  nitrogen  of  the  nitrite  therefore  has  been  oxidized 
from  3  to  5 — increased  2  steps.  The  chromium  of  the 
chromate  has  been  reduced  from  12  to  6 — decreased  6 


44  CHEMICAL   REACTIONS   AND   THEIR    EQUATIONS 

steps.  To  balance  the  factors  are  reversed,  6(2)  = 
2(6),  therefore, 

6NO2~  +  2Cr2O7—  =  6NO3-  +  4Cr+++ 

The  proportion  6  :  2  can  in  this  case  be  simplified  to  3  : 1 
and  the  unfinished  equation  becomes: 

3NO2-  +  O2O7—  =  3NO3-  +  2Cr+++ 
Second,  the  charges  are  counted  on  both  sides,  thus 

on  the  left  there  are  3  nitrites         =  3  negative  charges 

1  bichromate  =  2  negative  charges 

a  total  of          5  negative  charges 

on  the  right  there  are  3  nitrates       =    3  negative  charges 

2  chromic  ions  =    6  positive   charges 

a  total  of  3  positive  charges 

To  balance  +3  and  —5  requires  either  8  positive 
charges  to  the  left,  or  8  negative  charges  to  the  right. 
As  the  reaction  occurs  in  acid  solution,  eight  hydrogen 
ions  must  be  added  which  give  the  necessary  eight 
positive  charges  to  the  left  side,  and  the  equation  be- 
comes : 

3NO2-  +  Cr207-  +  8H+  =  3NO3-  +  2Cr+++ 
and  the  sum  on  both  sides  equals  3  positive  charges: 
3(-l)  +  (-2)  +  8(1)  =  3(-l)  +  2(3)  =  3 

Third,  the  atoms  are  balanced.  Nitrogen  and  chromium 
balance.  On  the  left  are  (3  X  2)  —  7  =  13  oxygen 
atoms,  while  on  the  right  there  are  only  (3  X  3)  =  9. 
This  excess  of  4  oxygen  atoms  at  the  left  will  give  with 
the  8  hydrogen  atoms  four  molecules  of  water  and  the 
finished  equation  becomes: 

(35)  3N02-  +  CraOr-  +  8H+  =  3N03-  +  2Cr+++4H20 


EQUATIONS  45 

In  a  few  cases  it  may  happen  that  a  substance  is 
oxidized  and  reduced  in  the  same  reaction,  that  is,  one 
part  of  the  substance  is  being  oxidized  and  the  other  is 
being  reduced.  It  is  then  advisable  to  write  the  formula 
of  this  substance  twice : 

KC1O3  +  KC103  =  KC1  +  KC1O4 

(+5)         (+5)        (-1)       (+7) 


This  diagram  shows  that  the  chlorine  (+5)  of  the 
chlorate  is  reduced  to  chloride  (  —  1),  while  another  part 
of  the  chlorate  is  oxidized  to  perchlorate  (7).  In  the 
first  case  the  valence  number  has  decreased  6  steps  and 
in  the  second  it  has  increased  2  steps.  It  follows,  that 
for  every  2  molecules  of  potassium  chlorate  giving  2 
molecules  of  potassium  chloride,  6  molecules  of  potas- 
sium chlorate  are  transformed  to  6  molecules  of  potas- 
sium perchlorate,  hence  8  molecules  of  potassium  chlorate 
give  2  molecules  of  potassium  chloride  and  6  mole- 
cules of  potassium  perchlorate,  which  can  be  simplified  to 

4KC1O3  =  KC1  +  3KC1O4 

QUESTIONS 

1.  Define  (a)  oxidation  and  reduction,  (b)  displacement,  (c)  stages  of 
oxidation,  (d)  ionic  equation,  and  (e)  non-ionic  equation. 

2.  What  is  the  cost  of  preparing  5  kg.  of  hydrogen  gas  from  zinc  and 
sulfuric  acid?     (Assume  the  price  of  Zn  to  be  50^  per  kilogram,  and  that 
of  H2SC>4  §i  per  kilogram.) 

3.  What  weight  of  phosphorus  will  completely  remove  the  oxygen 
from  2  kg.  of  air?     (Assume  that  air  contains  23  per  cent  of  oxygen 
by  weight.) 

4.  How  many  grams  of  oxygen  can  be  obtained  from  75  grams  of  a 
3  per  cent  solution  of  hydrogen  peroxide? 

5.  How  many  grams  of  a  6  per  cent  solution  of  hydrogen  peroxide  will 
convert  5  grams  of  lead  sulfide  into  -lead  sulf  ate  ? 

6.  How  many  grams  of  potassium  chlorate  are  necessary  to  furnish 


46  CHEMICAL    REACTIONS    AND    THEIR    EQUATIONS 

the  oxygen  for  conversion  of  25  grams  metallic  copper  into  cupric  oxide? 
(6)  How  many  grams  of  KC1  will  be  formed?  (c)  Can  this  reaction  be 
written  in  the  ionic  form?  (Give  reasons.) 

7.  How  many  grams  of  ammonium  nitrate  are  necessary  to -make  100 
grams  of  nitrous  oxide,  which  is  formed  by  heating  the  salt? 

8.  What  are  the  substances  formed  when  a  bromine  solution  is  added 
to  a  potassium  hydroxide  solution? 

9.  Write  an  equation  for  the  reaction  of   (a)   bromide,  manganese 
dioxide,  sulfuric  acid  to  give  free  bromine;  (6)  sodium  bromide,  manga- 
nese dioxide,  sulfuric  acid  to  give  free  bromine;  (c)  potassium  bromide, 
manganese  dioxide,  phosphoric  acid  to  give  bromine. 

10.  How  many  grams  of  50  per  cent  sulfuric  acid  can  be  made  from 
10  grams  of  sulfur  by  burning  the  sulfur  in  air,  and  passing  the  sulfur 
dioxide  thus  formed  over  finely  divided  platinum  which  acts  as  catalyser, 
thus  producing  sulfur  trioxide?     (Give  complete  set  of  equations.) 

11.  Twenty  kilograms  of  phosphate  rock,  Ca3(PO4)2,  are  heated  with 
sand,  SiOo,  and  coke,  C,  in  an  electric  furnace.     The  coke  burns  to  CO, 
the  phosphate  is  reduced  to  phosphorus,  and  CaSiO4  is  formed.     How 
many  grams  of  phosphorus  are  obtained  ? 

12.  Two  grams  of  an  alloy  of  Cu  and  Ag  are  dissolved  in  nitric  acid. 
To  this  solution  HC1  is  added.     The  precipitated  AgCl  weighs  1.8  grams, 
(a)  How  many  grams  of  Ag  and  Cu  were  present?     (b)  What  is  the  per- 
centage of  Ag  and  Cu  in  the  alloy  ? 

13.  Thirty  kilograms  of  an  ore  containing  60  per  cent  Sb2S3  are  heated 
with  iron.     How  many  kilograms  of  antimony  are  formed? 

14.  A  metric  ton  (=  1000  kg.)  of  pyrites  containing  8  per  cent  of 
foreign  substances  is  burned  in  a  sulfuric  acid  plant,     (a)  How   many 
kilograms  of  ferric  oxide  remain?     (6)  How  many  kilograms  of  sulfur 
dioxide  are  formed  ? 

15.  How  many  grams  of  (a)  bromine,   (b)  iodine  are   produced   by 
passing  a  current  of  chlorine  gas  through  a  solution  of  (a)  25  grams  of 
magnesium  bromide,  (b)  10  grams  of  potassium  iodide? 

16.  How  many  grams  of  (a)  water,  (b)  carbon  dioxide  are  formed  by 
burning  1  kg.  of  methane  (CH4)  ? 

17.  The  hydrogen  obtained  from  24.2  grams  of  zinc  and  hydrochloric 
acid  is  passed  over  heated  mercuric  oxide.     How  many  grams  of  water 
and  how  many  grams  of  mercury  are  produced? 

18.  How  many  grams  of  nitric  oxide,  NO,  are  formed  when  75  grams 
of  copper  are  dissolved  in  nitric  acid? 

19.  How  many  grams  of  chlorine  can  be  obtained  by  heating  43.5 
grams  of  manganese  dioxide  with  hydrochloric  acid? 

20.  How  many   grains    of  cryst.    copper  nitrate  must  be  heated  to 
redness  in  order  to  obtain  46.2  grams  of  NO2?     The  other  products 
formed  in  this  reaction  are  O2,  H2O,  and  CuO. 


EQUATIONS  47 

21.  How  many  grams  of  (a)  iron,  (6)  zinc,  (c)  aluminum  are  required 
to  precipitate  10.8  grams  of  silver  from  a  silver  nitrate  solution? 

22.  How  many  grams  of  iodine  will  be  precipitated  from  a  potassium 
iodide  solution  by  the  chlorine,  which  is  formed  from  21.75  grams  manga- 

•nese  dioxide  and  hydrochloric  acid? 

23.  How  many  kilograms  of  iron  and  sulfuric  acid  will  be  needed  to 
make  100  kg.  of  hydrogen  gas? 

24.  Balance  the  following  equations : 

(a)  Zn  +  Ag+  =  Zn++  +  Ag 

(b)  S—  +  C12  =  S  +  Cl- 

(c)  PC13  4-  KC103  =  POC13  +  KC1 

(d)  H3P02  =  PH3  +  H3P04 

(e)  Sb  +  Cl,  =  Sb2Cl3 
(/)  Sb205  =  Sb02  +  02 

(g)  KOC1  +  HOC1  =  KC1O3  -f  HC1 

(h)  NO  +  C12  =  NOC1 

(t)  CdSO4  +  CdS  =  CdO  +  SO2 

(j)  CrCl3  +  H2  =  CrCl2  +  HC1 

26.  Finish  and  balance  the  following  equations: 

(a)  HgO  +  C12  +  ?  =  HgCl2  -f  HOC1 

(b)  Bi  +  HN03  =  Bi(N03)3  +  ? 

(c)  Na  +  H2O  =  ?+  ? 

(d)  As  +  C12  =  ? 

(e)  (NH4)2Pt2Cl6  =  Pt  +  2C12  +  ? 
(/)  FeS  +  02  =  Fe203 

(g)  KOH  +  Cl,  =  KOC1  +  KC1  +  ? 
(W  KC1  +  F2  = 

(0-  Al  +  HC1  = 

(j)  CuO  -f  H2  = 
(A)  HgCl2  +  Hg  =  ? 

26.  Ordinary  gun  powder  is  a  mixture  of  saltpeter,  carbon,  and  sulfur. 
The  reaction  products  are  K2CO3,  K2SO4,  K2S4,  SO2,  and  N2. 

Complete  the  following  reactions : 

(a)  KNO3  -f  C  =  K2CO3  +  CO2  +  N2 

(b)  KN03  -f  C  +  S  =  K2S04  +  C02  -h  N2 

(c)  KNO3  +  C  +  S  =  K2S4  +  CO2  +  N2 

Assuming  that  these  are  all  the  reactions  concerned  in  an  explosion, 
calculate  the  correct  proportion  of  saltpeter,  carbon  and  sulfur. 

27.  Construct  equations  for  the  reduction  of  potassium  from  potassium 
sulfide  by  means  of  (a)  iron,  (b)  aluminum,  (c)  manganese.     The  reaction 
products  are  (a)  FeS,  (b)  A12S3,  and  (c)  MgS.     Which  metal  per  pound 
will  yield  the  largest  quantity  of  potassium?     Which  metal  is  most 
economical?     (A   pound    of    Fe   costs  5&  Al  =  25{,  and  Mg  =  30^.) 
Which  reaction  will  take  place  most  rapidly? 


CHAPTER  V 
CONTROL  OF  REACTIONS 

A  chemical  reaction  is  a  molecular  phenomenon  by 
which  the  composition  of  the  molecule  is  changed.  The 
piost  frequent  type  of  reactions  are  those  in  which  one 
part  of  the  molecule  is  exchanged  by  a  part  of  a  different 
molecule — thus  in  the  reaction  AB  +  CD  =  AD  +  CB 
we  may  say  that  B  has  changed  places  with  D,  or  A 
has  been  exchanged  by  B  for  C,  and  we  draw  the  infer- 
ence, that  either  A  has  a  greater  attractive  force  to  D 
than  to  B,  or  that  B  has  a  greater  attractive  force  to 
C  than  to  A.  Postponing  a  discussion  of  this  force  of 
attraction  holding  the  atoms  in  the  molecules  together, 
we  find  that  a  reaction  is  practically  never  100  per  cent 
efficient.  This  means  that  not  all  of  the  AB,  and  not 
all  of  the  CD  has  been  transformed  to  AD  and  CB.  In 
every  reaction  there  is  a  certain  and  definite  percentage 
of  transformation,  and  this  percentage  forms  the  basis 
of  the  measurements  in  physical  chemistry.  However, 
the  percentage  of  reacting  substances  and  reaction  prod- 
ucts is  not  the  only  factor  which  can  be  experimentally 
determined  for  a  given  reaction.  Every  reaction  requires 
a  certain  space  of  time;  some  may  be  instantane- 
ous ;  some  may  be  so  extremely  slow  as  to  be  hardly  recog- 
nizable. Ionic  reactions  or  reactions  which  occur  in 
solutions  are  usually  instantaneous,  while  molecular 
reactions  usually  require  a  measurable  period  of  time. 
For  both  kinds  of  reactions,  ionic  and  molecular,  it  is 
essential  that  the  ions  or  molecules  come  in  close  contact, 
for  if  the  two  ions  or  molecules  are  far  apart,  they  cannot 

48 


CONTROL  OP  REACTIONS  49 

react  upon  each  other.  The  contact  of  ions  or  molecules 
is  usually  termed  ionic  or  molecular  collision.  Any 
physical  means  which  facilitates  ionic  or  molecular 
collisions  will  thus  increase  the  possibility  of  a  reaction 
taking  place.  The  physical  means  which  increase  or  de- 
crease the  possibility  of  collisions  and  so  influence  a 
chemical  reaction  may  be  of  a  mechanical,  thermal,  or 
electrical  nature. 

Speed  of  Reaction. — The  speed  of  a  reaction  is  shown 
in  the  proportional  amount  of  reaction-products  (AD 
and  CB)  which  are  formed  from  the  reacting  substances 
( AB  and  CD)  in  a  given  period  of  time.  Thus  it  becomes 
evident  that  the  control  of  reactions  depends  entirely 
upon  the  regulation  of  the  speed  of  the  reaction.  By 
application  of  mechanical,  thermal,  or  electrical  means, 
either  alone  or  in  combination,  the  speed  of  a  reaction 
is  either  (a)  accelerated  (made  to  proceed  faster),  (6) 
retarded  (made  to  proceed  slower),  (c)  stopped  (made  to 
stand  still),  or  (d)  reversed  (made  to  proceed  in  the  op- 
posite direction).  Hence  for  every  reaction  there  are 
favorable  as  well  as  unfavorable  conditions. 

THE  MECHANICAL  CONTROL:  Surface,  Catalyzers,  Con- 
centration, and  Pressure 

Frequency  of  Collisions  is  Influenced  by  Number  of 
Molecules  Coming  in  Contact. — The  mechanical  means 
which  influence  the  frequency  of  molecular  collisions  are 
those  which  regulate  the  number  or  the  amount  of 
molecules.  The  more  molecules  collide  with  each  other, 
the  greater  the  possibility  of  a  reaction,  and  vice  versa. 
This  can  be  accomplished  in  four  different  ways:  (1)  by 
increasing  the  surface  of  contact  between  two  substances, 
(2)  by  mechanical  catalyzers,  (3)  by  higher  concentra- 
tion, (4)  by  greater  pressure. 


50  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

Increase  of  Surface  by  Subdivision. — An  increase  of 
the  contact  surface  between  two  substances  will  naturally 
increase  the  number  of  different  molecules  coming  into 
close  contact,  and  hence  increase  the  possibility  of  the 
molecules  colliding.  By  placing  a  cube  of  iron  contain- 
ing 1  c.c.  in  hydrochloric  acid,  it  is  evident  that  the 
contact  surface  between  the  solid  and  liquid  is  6  sq.  cm., 
for  the  edges  of  the  cube  are  each  1  cm.  long,  and  each 
side  has  a  surface  of  1  sq.  cm.,  and  there  are  six  sides  on 
the  cube.  If  this  cube  is  cut  into  eight  smaller  cubes,  as 
shown  in  Fig.  ] ,  the  volume  and  mass  of  the  iron,  as  well 


FIG.  1. — Increasing  the  surface  by  subdivision. 

as  the  total  number  of  iron  molecules  has  not  been 
changed,  but  the  surface  has  been  doubled,  making  it 
12  sq.  cm.,  and  consequently  twice  as  many  molecules 
of  iron  are  in  contact  with  hydrochloric  acid.  If  we 
divide  these  cubes  again  into  a  total  of  64  smaller  cubes, 
the  total  surface  of  the  same  amount  of  iron  will  be  23 
sq.  cm.  It  is  evident  that  this  division  can  be  theoretically 
continued  until  we  have  single  iron  molecules.  It  is  also 
evident  that  the  smaller  the  cubes,  the  greater  the  sur- 
face, and  the  greater  the  contact  surface  between  the 
solid  and  the  liquid,  therefore,  the  greater  the  possibility 
of  molecular  collisions.  It  is  thus  possible  to  predict  that 
the  single  large  iron  cube  will  dissolve  more  slowly  than 
the  four  small  ones,  and  that  the  64  smaller  ones  will  dis- 


CONTROL  OF  REACTIONS 


51 


solve  faster  than  the  four  small  ones.     The  following 
table  will  be  of  interest: 

INCREASING  THE  SURFACE  OF  A  1-c.c.  CUBE  BY  SUBDIVISION 


Edge  length 

Number 
of  cubes 

Total  surface 

Cubes  are 

Magnitude 

(a) 

(6) 

(c) 

(d) 

1  cm. 

1 

6  cm.  2 

visible 

coarse  granules 

0.  1  cm.  =  1  mm. 

103 

60  cm.  2 

visible 

fine  granules 

0.01  cm. 

10« 

600  cm.  2 

visible 

coarse  powder 

0.001  cm. 

10» 

6,000  cm.2 

microscopic 

fine  powder 

0.0001  cm.  =  In 

1012 

6  m.2 

microscopic 

very  fine  powder 

0.00001  cm. 

10" 

60m.2 

ultramicro- 

colloids 

0.000001  cm. 

10>8 

600  m.2 

scopic 

colloids 

0.0000001  cm.  =  IMM 

1021 

6,000  m.2 

amicroscopic 

colloids 

0.00000001  cm. 

1Q2« 

60,000  m.2 

or  invisible 

molecules 

(a)  1    m.    (meter)  =  100  cm.    (centimeter)  =  1,000  mm.    (millimeter)  =  1,000,000/i 
(microns)  =  1,000, 000, GOO^M  (millimicrons)  =  39.37  inches. 

(b)  103  =  1,000;  and  102*  =  1,000,000,000,000,000,000,000,000. 

(c)  Cm.2  =  cm.  X  cm.  =  square  centimeter;  m.2  =  m.  X  m.  =  square  meter. 

(rf)  Blood  corpuscles  7.5  to  15/i,  bacilli  1  to  10/u,  hydrogen  molecules  0.20/tM.  chlorine 
molecule  0.40/iM- 

Colloids. — Finely  divided  substances  usually  react 
faster  than  coarse  ones.  For  this  reason  finely  powdered 
substances  and  substances  in  the  colloidal  state  are 
reactive,  as  well  as  gases  and  substances  in  solutions. 
Colloids  are  simply  finely  divided  particles  of  ultramicro- 
scopical  size  which  form  the  transition  from  single 
molecules  to.  aggregations  of  molecules  of  microscopical 
size. 

Increase  of  Surface  by  Melting. — A  reaction  may  also 
be  started  when  the  surface  of  contact  between  two  solid 
substances  is  increased,  not  by  subdivision,  but  by  melt- 
ing one  substance.  Thus  in  Fig.  2  it  is  evident  that  the 
points  of  contact  between  two  substances  is  restricted 
when  both  are  solid  particles,  but  is  greatly  increased 
when  one  substance  is  liquid.  An  example  is  furnished 
by  finely  divided  sulfur  and  iron,  which  can  be  kept  mixed 


52 


CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 


together  indefinitely  without  a  reaction  occurring.  As 
soon,  however,  as  the  sulfur  is  molten,  the  reaction 
begins.  In  this  case  not  only  the  number  of  sulfur 
molecules  and  iron  molecules  colliding  with  each  other 
has  increased,  but  another  factor,  heat,  has  been  intro- 
duced, the  effect  of  which  forms  the  subject  of  a  later 
paragraph. 

Catalyzers. — A  reaction  may  also  be  either  accelerated 
or  retarded  by  the  presence  of  a  third  substance,  a 
catalyzer  or  catalyst,  which  is,  or  appears  to  be,  unchanged 
throughout  the  reaction  but  whose  presence  is  essential 
for  the  procedure  of  the  reaction.  The  action  of  a 


FIG.  2. — Increasing  the  surface  by  melting. 

catalyzer  is  either  of  a  mechanical  or  chemical  nature. 
The  mechanical  catalyzers  are  such  substances  which, 
when  finely  divided,  have  the  property  to  adsorb  upon 
their  surface  the  molecules  of  other  substances.  Thus, 
finely  divided  palladium  will  adsorb  hydrogen  gas  and 
the  surface  of  the  metal  will  be  covered  with  a  layer  of 
hydrogen  molecules.  These  densely  packed  hydrogen 
molecules  are  much  more  reactive  than  the  widely 
scattered  hydrogen  molecules  in  hydrogen  gas — in  other 
words  the  hydrogen  molecules  have  been  concentrated 
and  the  gas  has  been  condensed  to  a  solid  substance. 
Many  important  processes  in  industrial  manufacture 
depend  upon  the  use  of  catalyzers  (fixation  of  atmos- 
pheric nitrogen,  manufacturing  sulfuric  acid,  etc.). 


CONTROL  OF  REACTIONS 


53 


Mass  Action.  —  The  dependence  of  a  chemical  reaction 
on  the  concentration  of  the  molecules  is  expressed  in 
the  law  of  mass  action  by  Guldberg  and  Waage  which 
holds  that  the  chemical  effect  of  molecules  participating 
in  a  chemical  reaction  is  proportional  to  their  mass,  that 
is,  the  amount  or  number  of  molecules  present  in  a 
certain  volume.  In  the  case  of  solutions  this  mass  is 
expressed  as  concentration  which  is  the  amount  of 
substance  (gram-molecules  or  moles)  dissolved  in  a 
given  volume  of  solvent  (1  liter).  In  the  case  of  gases 
this  mass  is  expressed  as  pressure,  for  the  pressure  is  an 
indication  of  the  number  of  molecules  present  in  a 


FIG.  3. — Increasing  the  surface  by  concentration  or  pressure. 

certain  volume  of  a  gas.  This  is  best  illustrated  by 
Fig.  3  in  which  there  is  in  one  case  10  molecules  each 
in  a  given  volume,  and  in  the  second  case  20  molecules 
each  in  a  given  volume.  The  kinetic  theory  assumes 
that  the  molecules  of  a  gas  are  in  free  movement  and 
will  therefore  collide  and  bombard  the  sides  of  the 
vessels.  The  sum  of  these  impacts  upon  the  walls  of 
the  container  is  shown  as  pressure,  hence  by  doubling 
the  number  of  molecules  in  the  same  volume,  the  pres- 
sure is  doubled,  for  the  number  of  collisions  and  impacts 
has  been  doubled.  It  has  been  found  that  under  normal 
conditions  (atmospheric  pressure  of  760  mm.  at  0°C., 
and  at  sea-level)  one  mole  of  any  gas  will  occupy  a 


54  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

volume  of  22.4  liters.  (A  mole  is  the  molecular  weight 
of  a  substance  in  grams,  that  is,  a  mole  of  H2  is  2  grams, 
a  mole  of  O2  is  32  grams,  etc.)  The  kinetic  theory 
of  gases  has  been  extended  to  molecules  in  solutions 
which  means  that  the  molecules  and  ions  in  a  solution 
are  also  free  moving  and  exert  pressure — the  osmotic 
pressure. 

Pressure. — The  more  moles  there  are  in  a  liter  of  gas 
or  a  liter  of  solution,  the  more  molecules  there  are  in 
this  unit  volume,  and  the  higher  the  concentration  or 
pressure,  hence  the  greater  the  possibility  for  the  mole- 
cules to  collide  and  react.  The  tendency  to  react  is 
generally  increased  with  an  increase  in  the  number  of 
molecules  or  ions  present  in  a  certain  volume;  but  as, 
during  the  reaction,  some  of  these  molecules  or  ions 
have  been  used  up  or  exhausted  by  being  transformed, 
their  concentration  becomes  less  as  the  reaction  pro- 
ceeds, yet,  at  the  same  time  the  concentration  of  the 
reaction  products  is  increased  and  the  tendency  to 
again  form  the  original  molecules  becomes  greater. 
A  chemical  reaction  therefore  is  never  complete  if  the 
reaction  products  can  again  form  the  original  sub- 
stances. If  two  molecules  (AB  and  CD}  react  to  form 
two  other  molecules  (AD  and  CE)  as  reaction  products, 
then  after  a  certain  time  there  will  be  a  chemical  system 
in  which  all  four  types  of  molecules  (AB,  CD,  AD,  and 
and  CB)  are  present.  The  percentage  of  the  different 
molecules  will  depend  upon  the  speed  of  the  reactions 
AB  +  CD  ->AD  +  CB  and  AD  +  CB  -+AB  +  CD.  It 
is  evident  that  the  speed  of  the  first  reaction  is  greatest 
at  the  beginning,  for  the  reaction  is  started  with  a  high 
concentration  of  AB  and  CD,  and  the  absence  (zero 
concentration)  of  AD  and  CB,  but  this  speed  will  dimin- 
ish more  and  more  as  the  reaction  products  AD  and  CB 


CONTROL  OF  REACTIONS  55 

are  formed.  The  speed  of  the  second  reaction  is  at 
first  null,  as  no  AD  and  CB  is  present,  but  this  speed 
increases  slowly  with  an  increase  in  the  amount  or 
concentration  of  AD  and  CB.  As  the  speed  of  the 
first  reaction  diminishes  and  the  speed  of  the  second 
reaction  increases,  there  will  come  a  time  when  the  speed 
of  both  reactions  are  equal,  and  a  chemical  equilibrium 
is  established. 

Equilibrium. — In  a  chemical  equilibrium  the  visible 
effects  of  a  reaction  are  at  stand-still  and  the  reaction 
appears  to  have  stopped,  this  does  not  imply,  however, 
that  the  molecules  have  ceased  to  collide  with  each 
other.  The  molecular  collisions  still  take  place, 
but  the  formation  of  AB  and  CD  has  become  equal 
to  the  formation  of  AD  and  CB,  in  other  words  the 
exchange  of  the  constituents  of  the  molecules  is  balanced. 
The  same  condition  of  equilibrium  is  reached  when  the 
original  substances  are  AD  and  CB  instead  of  AB  and 
CD.  Under  definite  conditions  (temperature,  concen- 
tration or  pressure)  there  is  for  any  system  of  substances 
only  one  equilibrium  which  is  established  whether  the 
reaction  is  started  with  AB  and  CD  or  with  AD  and  CB. 

Example  of  Equilibrium. — A  concrete  example  is 
given  by  the  following  experiment.  In  one  closed 
vessel  hydrogen  gas  and  ferrous-ferric  oxide  is  heated, 
and  in  another  closed  vessel  iron  granules  and  steam 
(water).  After  heating  a  certain  time  and  keeping  at  a 
certain  temperature,  both  vessels  will  contain  the  same 
proportion  of  hydrogen  gas  to  steam.  In  the  first 
container  the  reaction  was 

(40)  4H2  +  Fe3O4  =  3Fe  +  4H2O 
while  in  the  second  vessel  the  reaction  was 

(41)  3Fe  +  4H2O  =  Fe3O4  +  4H2. 


56  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

Accordingly  one  reaction  is  the  reverse  of  the  other,  a 
fact  which  can  be  expressed  by  using  the  sign  of  a 
reversible  reaction,  namely  two  arrows: 

(42)  3Fe  +  4H2CMFe3O4  +  4H2 

In  both  cases  the  same  equilibrium  was  obtained,  though 
the  original  substances  were  different.  The  percentage 
of  steam  and  hydrogen  will  depend  upon  temperature, 
but  at  the  same  temperature  this  percentage  is  the  same 
in  both  cases. 

Shifting  the  Equilibrium. — The  experimental  condi- 
tions can  be  so  arranged  that  during  the  reaction  one 
or  the  other  of  the  gases  is  removed.  In  this  case  the 
removal  of  one  gas  will  constantly  shift  or  destroy  the 
equilibrium,  and  the  reaction  may  proceed  to  completion. 
Thus  by  heating  the  iron  granules  in  a  tube  open  at  one 
end,  and  passing  a  strong  current  of  steam  into  this 
tube,  hydrogen  gas  will  emerge  from  the  open  end  until  all 
the  iron  is  oxidized.  In  this  case  the  hydrogen  molecules, 
as  soon  as  formed,  are  carried  away  and  thus  removed 
by  the  excess  of  steam  flowing  into  the  tube,  and  the 
reaction  will  be  complete  in  one  direction: 

(43)  3Fe  +  4H20->Fe304  +  4H2 

The  concentration  of  hydrogen  molecules  under  these 
conditions  was  never  high  enough  to  initiate  the  re  verse 
reaction.  On  the  other  hand  if  the  tube  is  filled  with 
ferrous-ferric  oxide  and  a  stream  of  hydrogen  gas  is 
passed  into  it,  steam  will  be  formed  until  all  the  iron 
oxide  is  reduced  to  metallic  iron: 

(44)  Fe3O4  +  4H2->3Fe  +  4H2O 

In  this  case  likewise  the  reaction  product,  steam,  as 
soon  as  formed,  is  carried  away  by  the  current  pf  hydro- 


CONTROL  OF  REACTIONS  57 

gen  gas  and  the  reduction  of  the  iron  oxide  will  proceed 
completely. 

Speed  of  Reaction. — Theoretically,  the  speed  of  a 
chemical  reaction  is  greatest  in  the  beginning  but  di- 
minishes more  and  more  as  the  reaction  proceeds.  As  a 
rule  ionic  reactions  are  instantaneous,  that  is,  their 
speed  is  too  rapid  to  measure.  Molecular  reactions  are 
slower  and  their  speed  can  be  measured  and  controlled 
as  in  the  preceding  example.  Some  reactions  are  too 
slow  to  measure,  i.e.  the  oxidation  of  iron  in  air  at  ordi- 
nary temperature.  By  increase  of  concentration  (or 
temperature)  these  reactions  will  proceed  faster,  i.e. 
burning  of  iron  wire  in  oxygen. 

Ionic  Equilibrium. — The  control  of  an  ionic  reaction 
depends  upon  the  same  principle,  for  the  reaction  will 
take  place  whenever  the  ionic  equilibrium  is  disturbed 
or  shifted  by  changing  the  concentration  or  the  amount 
of  ions  present  in  a  unit  volume.  Assuming  for  example 
that  two  compounds,  AB  and  CD,  are  each  separately 
dissolved  in  certain  amounts  of  water  and  measure- 
ments show  that  the  first  solution  contains  90  per  cent 
ions  while  the  second  solution  contains  only  5  per  cent 
ions,  then  an  ionic  equilibrium  has  been  established 
which  can  be  expressed  by  the  equations: 

(a)    AB   ±^  A+  +  B-  and  (6)  CD  ±^  C+  +  D- 
10%      90%      90%  95%      5%        5% 

It  is  evident  that  in  solution  a  there  are  18  times  as  many 
ions  as  molecules,  hence  the  substance  AB  is  said -to  be 
highly  or  strongly  ionized;  while  in  solution  b  there  are 
nearly  ten  times  as  many  molecules  as  ions,  hence  the 
substance  CD  is  little  or  weakly  ionized.  "In  both  cases 
the  speed  of  the  reaction  from  left  to  right  and  from 
right  to  left  has  become  equal.  Such  an  ionic  equili- 


58  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

brium,  which  is  formed  whenever  a  substance  is  dis- 
solved in  water,  is  disturbed  by  either  adding  or 
removing  AB,  A+,  or  B~  to  or  from  solution  (a),  or  by  add- 
ing or  removing  CD,  C+,  or  D~  to  or  from  solution  (6). 
The  addition  or  removal  of  molecules  or  ions  is  accom- 
plished by:  (a)  adding  more  molecules  by  dissolving 
more  of  the  substance,  (6)  diluting  the  solution,  (c) 
removing  some  ions  in  the  form  of  molecular  or  non- 
ionized  soluble  compounds  (like  water,  weak  acids,  weak 
bases,  or  weak  salts),  (d)  removing  some  ions  in  an 
insoluble  compound  (precipitates),  (e)  removing  some 
ions  in  a  gaseous  compound  (gas  bubbles),  (/)  the  forma- 
tion of  complex  ions.  In  each  case  the  ionic  equilibrium 
is  shifted. 

Formation  of  Water. — Neutralization  means  the  add- 
ing of  H+  to  an  ionic  equilibrium  containing  OH~ 
whereby  water  is  formed,  and  as  water  is  practically  not 
ionized,  the  removal  of  H+  and  OH"  from  the  ionic 
equilibrium  will  be  practically  complete,  that  is  a  reac- 
tion will  take  place  until  all  H+  or  OH"  present  have 
been  used  up.  Therefore  whenever  there  is  the  possi- 
bility that  water  can  be  formed  from  two  solutions  a 
reaction  is  likely  to  take  place. 

Formation  of  Precipitate. — In  precipitation  a  mole- 
cular and  insoluble  compound  is  formed  from  ionic 
constituents  and  by  this  precipitate  certain  ions  are 
taken  from  the  solution  and  thereby  the  ionic  equilibrium 
is  shifted.  The  ionic  equilibrium  BaCl2  =  Ba++  + 
2C1"  can  be  made  to  proceed  from  left  to  right  by  either 
(a)  adding  more  barium  chloride,  (6)  removing  Ba++  as 
a  precipitate  (BaSO4)  by  the  addition  of  SO4 — ,  or  (c) 
removing  Cl~  as  a  precipitate  (AgCl)  by  adding  Ag+. 
In  the  last  two  cases  the  ions  have  been  removed  by  a 
precipitate  hence  more  of  the  molecular  barium  chloride 


CONTROL  OF  REACTIONS  59 

will  ionize,  and  if  sufficient  SO4"~  or  Ag+  is  added,  all 
of  the  ions  may  be  precipitated,  and  thereby  the  reaction 
controlled. 

Formation  of  Gas. — An  ionic  equilibrium  is  likewise 
shifted  by  the  formation  of  a  gas  from  certain  ions,  and 
the  liberation  of  the  gas  will  remove  certain  ions  from 
the  solution.  Thus,  by  adding  an  acid  (H+)  to  a  solution 
of  sodium  sulfide  Na2S  =  2Na+  +  S~ ,  the  formation  of 
gaseous,  non-ionized  hydrogen  sulfide  will  remove  the 
S —  from  the  equilibrium,  hence  more  Na2S  will  ionize 
and,  if  sufficient  acid  is  added,  all  of  the  sulfide  ions  may 
be  removed  as  H2S.  Such  reactions  can  be  accelerated 
by  mechanical  means  (e.g.  agitation  or  passing  a  current 
of  air  through  the  solution)  or  thermal  means  (e.g. 
heating)  or  both  together,  for  by  these  means  the  escape 
or  liberation  of  the  gas  from  the  solution  is  facilitated. 

Formation  of  Weak  Acids. — The  formation  of  weak 
acids  (acids  which  are  little  ionized)  will  remove  hydrogen 
ions.  A  solution  of  sulfuric  acid  contains  much  H+, 
that  is, 

H2SO4  2H+         +       S04- 

(small  %)  (large  %)  (large  %) 

and  when  this  is  added  to  a  soluble  acetate,  citrate,  oxalate, 
or  other  salts  of  a  weak  acid,  the  acidity  or  hydrogen  ion 
concentration  will  be  reduced,  as  the  respective  slightly 
ionized  acetic  acid,  citric  acid,  or  oxalic  acid  is  formed: 

(45)  H+  +  Ac-  =  HAc 

To  illustrate  this  by  experiment,  fill  a  test  tube  with 
very  dilute  sulfuric  acid  and  color  it  with  an  indicator 
(litmus,  methyl  orange,  methyl  violet,  congo  red,  etc.), 
if  the  neutral  salt  (e.g.  sodium  acetate,  sodium  citrate) 
is  added,  the  color  will  change  and  indicate  a  neutral 
reaction. 


60  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

Formation  of  Weak  Bases. — The  formation  of  weak 
bases  (bases  which  are  little  ionized)  will  remove  hydroxyl 
ions.  Thus  when  a  solution  of  sodium  hydroxide  (strong 
base)  containing  a  large  percentage  of  OH~,  is  added  to 
a  solid  ammonium  salt,  the  reaction  product  will  be  am- 
monium hydroxide  (weak  base)  which  is  but  little  ionized : 

(46)  OH-  +  NH4+  =  NH4OH 

Try  the  following  experiment:  Color  a  diluted  sodium 
hydroxide  solution  with  a  few  drops  of  indicator  and 
add  some  ammonium  sulfate  or  ammonium  chloride- 
note  the  change  in  color. 

Formation  of  Weak  Salt  and  Complex  Ion. — A  weak 
salt  is  a  compound  which,  while  soluble,  is  little  ionized, 
MN*=?M+  +  N~,  with  a  large  percentage  of  MN,  and  a 
very  small  percentage  of  M+  and  N~.  If  such  a  com- 
pound is  formed  during  a  reaction  its  formation  will 
remove  certain  anions  and  cations  and  thereby  destroy 
the  ionic  equilibrium.  Likewise  the  formation  of  com- 
plex ions  destroys  the  ionic  equilibrium  by  the  removal 
of  ions.  A  complex  ion  is  a  charged  group  of  atoms 
or  an  ion  combined  with  a  neutral  molecule,  such 
as  the  ammonia  complexes  Cu(NH3)4++,  Ni(NH3)4++, 
Ag(NH3)2+,  etc.;  the  cyanide  complexes  Cu(CN)2+'f, 
Ni(CN)4++,  etc.;  certain  halogen  compounds,  oxalates, 
ferro-  and  ferricyanides,  etc. 

Prediction  of  Reaction. — To  summarize  the  facts  of 
ionic  equilibrium  and  ionic  concentration:  A  reaction  is 
likely  to  occur  whenever  there  is  a  possibility  that  the 
ions  of  two  different  molecules  in  solution  (ionic  equilib- 
rium) may  form  (a)  water,  (6)  precipitate,  (c)  gas,  (d) 
weak  acids,  (e)  weak  base,  (/)  weak  salt,  or  (g)  complex 
ions,  for  by  the  formation  of  these  substances  the  ionic 
concentration  is  changed  (decreased).  By  applying 


CONTROL  OF  REACTIONS  61 

this  rule  it  can  be  predicted  whether  or  not  a  reaction 
will  occur  if  two  substances  in  solution  are  brought 
together.1 

THE  THERMAL  CONTROL 

Increase  or  Decrease  of  Temperature. — The  thermal 
means  which  influence  the  frequency  of  molecular  collisions 
are  those  by  which  not  the  number  but  the  velocity  of  the 
molecules  is  regulated.  The  faster  the  molecules  move 
the  more  numerous  are  the  collisions;  the  slower  their 
movement,  the  less  becomes  the  frequency  of  collisions. 
Heat  or  a  rise  in  temperature  increases  the  molecular 
movement,  hence  increases  the  nnmber  of  molecular  collis- 
ions and  therefore  accelerates  a  reaction.  A  rise  of  10°C. 
will  generally  double  the  speed  of  a  reaction. 

States  of  Aggregation. — It  is  assumed  that  the  three 
states  of  aggregations,  solids,  liquids,  and  gases,  are  due 
to  the  rate  of  vibrations  of  the  molecules.  In  solid  and 
liquid  substances  the  vibrations  are  comparatively  slow, 
while  in  dissolved  and  gaseous  substances  the  vibrations 
are  fast  and  the  molecules  can  move  freely.  The 
change  from  the  solid  to  the  liquid  state,  and  from  the 
liquid  to  the  gaseous  state  is  thus  always  a  change  from 
slower  to  faster  vibrating  molecules.  Any  change  in 
the  velocity  or  vibrating  speed  of  the  molecules  manifests 
itself  either  in  an  absorption  or  liberation  of  heat.  We 
speak  therefore  of  heat  of  fusion  or  melting  as  well  as 
heat  of  vaporization  and  mean  thereby  the  amount  of 
heat  necessary  to  transform  a  solid  at  its  melting  point 
into  a  liquid,  or  a  liquid  at  its  boiling  point  into  a  gas. 

Exothermic  and  Endothermic. — During  a  chemical 
reaction  heat  may  be  produced  or  consumed.  If  the 

1  For  a  full  discussion  of  the  control  of  chemical  reactions  and  equilib- 
rium see  Hildebrand's  Principles  of  Chemistry,  Chaps,  xi-xiv. 


62  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

reaction  liberates  heat  (exothermic  reaction)  then  a 
compound  (exothermic  compound)  has  been  formed  ^as 
reaction  product  which  contains  less  heat  than  the  previ- 
ous molecules,  for  the  molecules  of  one  or  the  other 
reaction  product  will  vibrate  at  a  lower  speed  than  the 
original  reacting  substances.  Such  molecules  of  exo- 
thermic compounds  are,  as  a  rule,  stable  and  not  readily 
decomposed.  If,  on  the  other  hand,  during  a  reaction  a 
certain  amount  of  heat  is  used  up  (endothermic  reaction) 
then  a  compound  (endothermic  compound)  has  been 
formed  whose  molecules  vibrate  more  rapidly.  Endo- 
thermic compounds  are,  as  a  rule,  unstable  and  may 
decompose  so  rapidly  as  to  cause  explosions.  The 
relation  between  these  two  types  of  reactions  and  com- 
pounds is  shown  in  the  sketch: 

EXOTHERMIC  REACTION 

(proceeds  rapidly  from  left  to  right  and  liberates  heat) 
AB  +      CD       <=±          AD         +        CB 

ENDOTHERMIC  COMPOUNDS    EXOTHERMIC  COMPOUNDS 
(unstable)  (stable) 

ENDOTHERMIC  REACTION 
(proceeds  slowly  from  right  to  left  and  absorbs  heat). 

An  exothermic  reaction  is  the  change  from  endothermic 
to  exothermic  compounds — from  faster  to  slower  vibrat- 
ing molecules.  This  change  requires  no  heat,  only  an 
external  start  and  the  reaction  will  proceed  rapidly 
and  completely  with  the  liberation  of  heat,  it  may  be 
even  explosive.  An  endothermic  reaction  is  the  change 
from  an  exothermic  to  endothermic  compound, — from 
slower  to  faster  vibrating  molecules.  This  change 


CONTROL  OF  REACTIONS  63 

requires  a  constant  influx  of  heat  and  the  reaction 
will  proceed  slowly  and  be  limited  by  the  tendency  of 
the  reaction  products  to  again  decompose  (dissociation). 

The  general  tendency  of  chemical  reactions  is  to 
produce  exothermic  compounds,  that  is,  reaction  products 
which  develop  the  largest  amount  of  heat  (Thomson's  rule, 
Berthelot's  principle).  In  other  words,  rapidly  vibrating 
molecules  have  the  tendency  to  rearrange  themselves  and 
form  slower  vibrating  molecules,  thereby  liberating  a  defi- 
nite amount  of  heat.  However  if  external  heat  is  applied 
to  two  exothermic  compounds,  their  molecular  vibrations 
may  be  accelerated  sufficiently  to  form  an  endothermic 
compound. 

Control  of  Velocity. — The  rise  or  fall  of  temperature 
therefore  is  a  means  to  increase  or  decrease  the  velocity 
of  molecular  vibrations  which  in  turn  controls  the  speed 
of  molecular  reactions.  The  application  of  heat 
generally  has  no  effect  upon  ionic  reactions. 

It  is  also  evident  that  the  increasing  velocity  of 
molecular  vibrations  can  aid  combination  as  well  as 
decomposition,  or  dissociation.  Dissociation  or  the 
breaking  apart  of  molecules  may  be  caused  by  the  two 
external  means — lieat  and  electricity.  The  thermic 
dissociation  (commonly  called  dissociation)  is  produced 
by  higher  temperatures  and  continues  only  as  long  as 
the  external  cause  (heat)  is  active;  the  electric  dis- 
sociation (properly  called  ionization)  is  produced  by 
dissolving  electrolytes. 

Dissociation. — Thermic  dissociation  proceeds  slowly 
with  the  rise  of  temperature  and  diminishes  accordingly 
with  a  fall  of  temperature.  I.e.  hydrogen  and  oxygen 
begin  to  combine  at  200°C.  and  form  steam;  at  1,200°C. 
dissociation  begins,  for  some  molecules  of  H2O  will 
decompose  into  hydrogen  and  oxygen;  at  2,500°C. 


64  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

there  is  about  half  of  the  steam  dissociated  into  H2  and 
O2,  and  with  a  further  rise  of  temperature  all  the  mole- 
cules of  steam  are  finally  dissociated.  By  lowering  the 
temperature  combination  takes  place  again  and  at  1,200° 
C.  only  steam  will  be  present. 

Dissociation  does  not  depend  solely  upon  temperature, 
but  also  upon  pressure.  Increase  in  pressure  diminishes, 
while  a  decrease  in  pressure  increases  dissociation.  For 
each  definite  pressure  at  a  definite  temperature  the 
amount  of  dissociation,  that  is,  the  percentage  of  dis- 
sociated molecules  of  the  same  substance  will  be  con- 
stant. The  dissociation  may  be  diminished  by  adding 
more  of  one  of  the  gaseous  components,  for  then  the 
equilibrium  is  shifted  by  the  introduction  of  a  larger 
amount  of  one  gas.  Dissociation  is  not  restricted  to 
gaseous  compounds  and  another  example  that  heat  can 
combine  as  well  as  decompose  is  the  classical  experiment 
of  Lavoisier  of  making  oxygen  from  mercuric  oxide. 
At  ordinary  temperature  mercury  is  not  attacked  by  the 
oxygen  of  the  air,  but  heating  slowly  transforms  it  into 
mercuric  oxide,  and  if  heated  to  a  higher  temperature, 
it  decomposes  again  into  mercury  vapor  and  oxygen. 

The  explanation  of  dissociation  is  found  in  the  assump- 
tion that  not  only  the  molecules  but  also  the  atoms  are 
in  movement  or  rather  vibrating.  Therefore  a  rise  of 
temperature  will  not  only  increase  the  speed  of  the 
molecules,  but  will  also  increase  the  speed  of  the  vibra- 
tions of  the  atoms,  and  finally  the  atomic  vibrations  will 
become  so  rapid  that  the  atoms  will  leave  their  attraction 
spheres  and  travel  as  independent  units — thus  splitting 
the  molecules. 

Action  of  Light. — The  action  of  light  is  probably  of  a 
similar  nature.  Light  rays  either  combine  or  decompose. 
Thus  a  mixture  of  C12  and  H2  when  kept  in  the  dark 


CONTROL  OF  REACTIONS  65 

remains  unchanged,  but  if  brought  into  bright  sunlight, 
the  gases  may  combine  so  suddenly  that  the  mixture 
explodes.  The  action  of  light  upon  the  photographic 
plate  and  the  bleaching  of  aniline  dyes  are  examples  of 
decomposition  by  light. 

THE  ELECTRICAL  CONTROL:  Electro-motive  Force 

The  electrical  means  which  influence  a  chemical  reac- 
tion depend  upon  the  control  of  the  electro-motive  force 
or  potential  energy.  The  relationship  between  chemical 
and  electrical  energy  is  close  and  there  are  various  ways 
to  transform  one  into  the  other  and  vice  versa. 

Electrolysis. — The  transformation  of  electrical  into 
chemical  energy  is  illustrated  by  the  phenomena  of 
electrolysis  and  displacement.  When  an  electric  current 
is  passed  through  a  molten  or  dissolved  electrolyte,  that 
is,  any  substance  which  conducts  the  current,  separation 
or  electrolysis  takes  place.  The  terms  employed  in 
electrolysis  are  illustrated  in  the  sketch: 

ELECTRODES 
or  poles 


Positive  electrode  Negative  electrode 

ANODE  CATHODE 

IONS 
or  charged  atoms 


negative  ion        positive  ion 
ANION  CATION 

< — e  © — > 

Electrolysis  can  take  place  only  when  the  electrolyte  is 
either  molten  or  dissolved,  that  is,  already  more  or  less 
dissociated  or  split  into  ions;  hence  it  is  not  a  decomposi- 


66  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

tion  of  the  electrolyte  but  merely  a  separation  of  the 
ions  already  present.  The  cause  of  this  separation  or 
migration  of  the  ions  to  the  poles  or  electrodes  is  found 
in  the  charge  of  the  atoms.  The  anions  are  negatively 
charged  atoms  and  will  migrate  or  move  to  the  anode  or 
positive  pole,  while  the  cations  or  negatively  charged 
atoms  move  toward  the  cathode  or  negatively  charged 
electrode  and  -there  deliver  their  charges.  The  charges 
of  the  ions  are  taken  up  by  the  electrodes  or  poles  and 
the  ions  which  lose  their  charge  are  then  deposited  as 
atoms  or  molecules  of  the  free  elements. 

In  the  electrolysis  of  any  acid  the  hydrogen  is  found 
at  the  cathode,  and  the  acid  radical  at  the  anode.  In 
the  electrolysis  of  any  base  the  metal  is  liberated  at  the 
cathode,  while  the  hydroxyl  group  moves  to  the  anode. 
In  the  electrolysis  of  any  salt  its  metallic  constituent  is 
found  at  the  cathode,  while  the  non-metal  or  acid  radical 
is  deposited  at  the  anode.  Thus  hydrogen  and  metals 
are  considered  positive  elements,  while  non-metals  and 
acid  radicals  are  negative  elements  or  radicals.  The 
atoms  or  group  of  atoms  deposited  at  the  electrodes  and 
deprived  of  their  charges  can  not  exist  in  free  state, 
hence  will  combine  and  form  molecules. 

Secondary  Reaction  Products. — In  many  cases  these 
newly  formed  molecules  react  either  with  the  material 
of  the  electrode  or  with  the  solvent  producing  new  com- 
pounds as  secondary  reaction  products.  In  such  cases, 
not  the  elements  of  the  electrolyte,  but  secondary  reac- 
tion products  are  found.  If  an  electric  current  is 
passed  through  a  solution  of  sodium  chloride  (electro- 
lyte), the  sodium  (cation)  migrates  to  the  negative 
electrode  (cathode),  and  the  chlorine  (anion)  moves  to 
the  positive  electrode  (anode)  and  forms  respectively 
sodium  metal  and  chlorine  gas,  but  these  free  elements 


CONTROL  OF  REACTIONS  67 

both  react  with  water  and  thus  give  sodium  hydroxide, 
at  the  cathode  and  at  the  anode  hydrochloric  acid 
and  hypochloric  acid.  These  important  secondary  re- 
action products  account  for  many  phenomena  and 
their  formation  is  utilized  in  many  industrial  processes 
for  the  manufacture  of  chemicals.  Formerly  it  was 
believed  that  the  electric  current  electrolyzed  or  de- 
composed water  into  hydrogen  and  oxygen  gas.  Now 
we  know  that  absolutely  pure  water  is  a  very  poor 
electrolyte  and  will  not  conduct  the  electric  current, 
hence  it  cannot  be  decomposed.  However  if  an  electro- 
lyte is  added,  i.e.,  if  the  water  is  acidulated  with  sulfuric 
acid,  then  these  two  gases  will  be  produced,  not  as 
primary,  but  as  the  secondary  reaction  products.  What 
really  happens  is  the  decomposition  of  the  sulfuric  acid 
which  is  dissociated  or  ionized  into  the  cations  H+  and 
anions  SO4~~.  The  electric  current  simply  separates 
these  ions  locally  and  the  H+  of  the  sulfuric  acid  delivers 
its  charge  at  the  cathode  and  forms  H2  molecules; 
while  the  anions  SO4~  ~  migrate  to  the  anode,  deliver 
their  charges,  and  react  with  water  to  form  sulfuric  acid 
and  oxygen: 

4H+  =  2H2  +  4(+)  at  cathode, 
2SO4-    +  2H2O  =  2H2SO4  +  O2  +  4(-)  at  anode 

Combining  these  two  reactions  and  cancelling  the  sul- 
furic acid  there  remains: 

(47)  2H20  =  2H2  +  02 

which  is  the  visible  part  of  this  electrolysis,  for  when  the 
gases  are  collected  the  volume  of  hydrogen  will  be 
exactly  twice  the  volume  of  oxygen.  The  sulfuric  acid 
formed  at  the  anode  dissociates  again : 

(48)  2H2SO4  =  4H+  +  2SO4— 


68  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

and  its  respective  cations  and  anions  migrate  once  more 
to  the  cathode  and  anode  and  repeat  the  process.  In 
the  electrolysis  of  acidulated  water  there  is  thus  a 
continuous  dissociation  and  combination  of  the  acid  in 
which  the  constituents  of  water  take  part,  and  by  which 
hydrogen  and  oxygen  are  formed  as  secondary  reaction 
products. 

Electrochemical  Equivalent.— The  amount  of  substance 
deposited  or  liberated  at  the  electrodes  depends  upon 
the  amount  of  electricity  which  passes  through  the 
solution.  The  amount  of  electricity  is  measured  in 
coulombs  or  ampere-seconds.  The  same  quantity  of 
electricity  in  the  same  length  of  time  will  separate  the 
ions  at  both  electrodes  in  proportion  to  their  equivalent 
weight  (Faraday's  rule).  The  equivalent  weight  is  the 
atomic  weight  divided  by  the  valency.  To  deposit  a 
gram  equivalent  (atomic  weight /valence  =  x  grammes) 
of  any  ion  requires  96,540  coulombs.  Hence  of  copper 
(Cu++)  63,6/2  =  31.8  grams,  of  silver  (Ag+)  107.0 
grams,  of  ferrous  ion  (Fe++)  56/2  =  28  grams,  of  ferric 
ion  (Fe+++)  56/3  =  18.7  grams  will  be  deposited  by 
96,540  coulombs.  The  total  electrical  charge  on  a  gram 
equivalent  of  any  ion  is  therefore  96,540  coulombs  as 
this  amount  of  electricity  is  required  to  neutralize  the 
total  charge  of  the  gram  equivalent;  and  the  total 
charge  on  a  mole  or  gram  atom  of  an  ion  is  96,540  times 
its  valency. 

Electro-motive  Force. — The  force  with  which  the 
ions  hold  this  charge  is  called  electro-affinity  or  electro- 
motive force.  This  force  varies  for  different  ions.  To 
the  strong  ions  belong  the  cations  K+,  Na+,  Li+,  and 
the  anions  NO3~,  F~,  Cl~,  SO4 — ;  while  to  the  weak  ions 
belong  the  cations  Hg++,  AU+++,  Pt++++,  and  the  anions 
O — ,  S — ,  CN~.  (It  is  well  to  note  the  distinction 


CONTROL  OF  REACTIONS  69 

between  the  ionic  charge  and  the  force  with  which  the 
ionic  charge  is  held.) 

A  substance  of  strong  electro-motive  force  is  one  which 
produces  strong  ions,  that  is,  the  atoms  hold  the  electrical 
charge  firmly.  A  substance  of  weak  electro-motive 
force  is  one  whose  atoms  hold  the  electrical  charge  less 
firmly,  and  thus  gives  weak  ions. 

Displacement. — If  a  substance  capable  of  yielding 
strong  ions  be  added  to  a  solution  of  weak  ions,  then  the 
weak  ions  will  give  up  their  charge  and  become  electri- 
cally neutral,  while  the  solution  will  contain  the  strong 
ions.  When  a  piece  of  metallic  zinc  (strong  electro- 
motive force)  is  placed  in  a  solution  containing  lead 
ions  (weak  electro-motive  force),  zinc  ions  are  formed, 
and  metallic  lead  is  deposited  as  a  black  spongy  mass : 

(49)  Zn  +  Pb++  =  Zn++  +  Pb 

Metallic  lead  in  turn  will  precipitate  copper  (weaker 
electro-motive  force)  from  a  solution  of  cupric  ions: 

(50)  Pb  +  Cu++  =  Pb++  +  Cu 

Metallic  copper  will  discharge  mercuric  ions  and  deposit 
mercury : 

(51)  Cu  +  Hg++  =  Cu++  +  Hg 

The  weak  electro-motive  force  of  mercury  is  stronger 
than  the  electro-motive  force  of  silver,  therefore  it  will 
take  the  ionic  charge  from  the  silver  ions  and  crystalline 
silver  will  be  deposited: 

(52)  Hg  +  2Ag+  =  Hg++  +  2Ag 

In  the  same  manner  electro-negative  elements  of 
stronger  electro-motive  force  will  become  ionized  and 
form  anions  if  they  come  in  contact  with  anions  of 


70  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

weaker  electro-motive  force,  hence  chlorine  gas  will 
react  with  bromine  ions: 

(53)  C12  +  2Br-  =  2C1~  +  Br2 
Bromine  will  react  with  iodide  ions: 

(54)  Br2  +  21-  =  2Br-  +  I2 
Iodine  is  stronger  than  sulftde  ion : 

(55)  I2  +  S—  =  21-  +  S 

Displacement  Series. — The  sequence  in  which  the 
elements  act  in  this  manner  is  according  to  the  preceding 
equations  for  the  positive  metals  Zn  >  Pb  >  Cu  >  Hg 
>Ag,  and  for  the  negative  non-metals  Cl  >  Br>  I >S. 
Zinc  is  thus  more  positive  than  silver,  and  silver  more 
negative  than  zinc.  Chlorine  is  more  negative  than 
sulfur,  or  sulfur  is  more  positive  than  chlorine.  Positive 
and  negative  are  thus  relative  terms  and  it  is  possible 
to  place  all  elements  in  such  a  relative  series  which  is 
termed  the  displacement  series.  In  the  displacement 
series,  given  in  the  appendix,  the  most  negative  elements 
are  at  the  beginning,  and  these  will  displace  as  anions  all 
that  follow;  while  the  elements  at  the  end  are  the  most 
positive  and  will  displace  as  cations  all  above  them. 

From  the  above  it  becomes  evident  that,  if  a  piece  of 
iron  is  dipped  into  a  copper  salt  solution,  metallic  copper 
will  be  deposited  and  the  iron  will  go  into  solution: 

(56)  Fe  +  Cu++  =  Fe++  +  Cu 

The  cupric  ion  has  given  its  charges  to  the  iron  and  has 
become  electrically  neutral  copper,  while  the  previously 
electrically  neutral  iron  has  taken  this  charge  and  become 
ferrous  ion.  This  transmission  of  the  electric  charge 
takes  place  simultaneously  and  may  occur  at  any  point 
of  contact  between  the  iron  and  the  copper  ions. 


CONTROL  OF  REACTIONS 


71 


Voltaic  Cells. — The  transformation  of  chemical  energy 
into  electrical  energy  is  demonstrated  by  voltaic  cells  or 
batteries.  Thus  the  experimental  conditions  of  the  above 
reaction  (56)  can  be  so  arranged  that  the  direct  contact  of 
iron  and  copper  ions  is  prevented,  but  the  exchange  of  the 
charges  or  electrons  is  made  possible  by  a  conducting 
connection  between  the  iron  and  the  copper  ions  which 
causes  a  stream  of  electron  to  pass,  thus  producing  an 
electric  current  in  the  conducting  connection  (metal 


FIG.  4. — A  voltaic  cell  or  electrical  battery. 

wire).  These  experimental  conditions  are  illustrated  in 
a  galvanic  battery  or  voltaic  cell.  It  is  essential  in 
such  experimental  conditions  that  the  reacting  sub- 
stances are  locally  separated,  yet  connected  by  a  con- 
ducting media,  for  only  then  are  the  electrons  forced  to 
travel  through  the  conductor  and  produce  an  electric 
current  when  the  circuit  is  closed.  The  conditions  are 
schematically  represented  in  Fig.  4. 

The  farther  apart  the  two  metals  selected  as  electrodes 
stand  in  the  displacement  series,  the  greater  will  be  the 
produced  electro-motive  force  or  potential,  measured  in 


72  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

volts.  Hence  Fe  and  Zn  will  produce  a  smaller  voltage 
than  Al  and  Pb  or  Mg  and  Cu;  likewise  Zn  and  Cu 
or  Zn  and  Ag  will  produce  a  greater  voltage  than  Zn 
and  Fe. 

Example  of  Cell. — In  the  figure  a  rod  of  iron  and  a  rod 
of  copper  are  used  as  electrodes.  The  iron  stands  in 
a  solution  of  ferrous  sulfate,  -the  copper  in  a  solution  of 
cupric  sulfate,  and  the  solutions  are  separated  by  a 
porous  wall  or  semi-permeable  membrane.  If  the 
concentration  of  the  ferrous  sulfate  solution  is  low,  and 
the  concentration  of  the  copper  sulfate  solution  is  high— 
the  metallic  iron  tends  to  go  into  solution,  that  is,  acquire 
two  positive  charges  and  thus  form  ferrous  ions,  while, 
at  the  same  moment,  a  cupric  ion  will  deliver  two  positive 
charges  to  the  copper  rod  and  be  deposited  as  metallic 
copper.  The  iron  rod  will  lose  weight;  the  copper  rod 
will  gain  weight.  The  iron  electrode  becomes .  "  less 
positive"  that  is  "more  negative,"  and  the  copper 
electrode  becomes  more  positive.  The  reactions  which 
take  place  are : 

at  the  cathode  (57)  Fe  +  2(  +  )  =  Fe++ 
at  the  anode    (58)  Cu++  =  Cu  +  2(+) 

and  by  adding  these  two  equations,  and  cancelling  the 
two  positive  charges  which  appear  on  both  sides,  equa- 
tion (56)  results. 

Both  reactions  (57  and  58)  occur  simultaneously 
and  proceed  until  an  equilibrium  is  established, 
and  the  current  ceases.  If  the  ferrous  ion  concentra- 
tion is  very  low  in  the  beginning  and  the  cupric  ion 
concentration  very  high,  the  battery  will  live  longer. 
If,  however,  at  the  beginning  the  ferrous  ion  concentra- 
tion is  high,  and  the  cupric  ion  concentration  is  low, 
little  or  no  current  will  be  produced. 


CONTROL  OF  REACTIONS  73 

Secondary  Cells. — By  the  proper  selection  of  the 
materials  used  as  electrodes  and  solutions,  many  of  the 
batteries  can  be  arranged  so  that  the  reaction  is  reversi- 
ble. In  such  a  case  an  electric  current  passed  through 
the  cell  in  the  reverse  direction  will  cause  the  reverse 
reaction  and  the  original  state  will  be  resumed.  Such 
would  be  the  case  in  Fig.  4  if  the  copper  were  connected 
with  the  negative  pole  of  a  dynamo,  and  the  iron  with 
the  positive  pole  so  that  electrolysis  would  take  place. 
Such  a  battery  is  called  a  secondary  cell  or  accumulator. 

Accumulator. — At  present  the  lead  accumulator  is 
the  most  practical  secondary  cell  for  common  use. 
In  discharging  this  accumulator,  the  lead  is  the  cathode, 
the  lead  oxide  the  anode,  and  the  following  reaction 
takes  place: 

at  the  cathode  (59)  Pb  +  2(+)  =  Pb++ 
at  the  anode    (60)  PbO2  +  2(-)  =  PbO2~ 

Thus  at  the  cathode  the  lead  will  dissolve  and  send 
positive  lead  ions  into  solution,  and  the  cathode  becomes 
less  positive — hence  negative.  From  the  anode  negatively 
charged  PbO2 —  ions  go  into  solution,  and  the  anode  thus 
loses  negative  charges  or  electrons,  it  becomes  less 
negative, — hence  more  positive.  As  there  is  an  excess 
of  sulfuric  acid  in  the  accumulators  the  lead  ions  form 
insoluble  lead  sulfate: 

(61)  Pb++  +  S04—  =  PbS04 

The  plumbate  ions  formed  at  the  anode  react  with  the 
hydrogen  ions  of  the  sulfuric  acid  and  give  plumbic 
acid,  which  in  turn  reacts  with  the  sulfuric  acid  and 
yields  ultimately  lead  sulfate. 

(62)  PbO2-  +  2H+  =  H2PbO2 

(63)  H2PbO2  +  2H+  +  SO4—  =  PbSO4  +  2H2O 


74  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

The  endproduct  in  both  cases  is  thus  lead  sulfate  and 
theoretically  the  accumulator  is  entirely  exhausted  or 
discharged  when  all  the  lead  arid  lead  oxide  has  been 
transformed  into  lead  sulfate;  but  practically  this 
condition  never  happens  as  the  reaction  is  reversible. 
When  an  accumulator  is  exhausted  a  chemical  equilib- 
rium is  established  in  which  the  reactions  59  to  63  have 
ceased  and  a  large  amount  of  lead  sulfate  is  present. 
By  passing  an  electric  current  from  an  outside  source 
through  the  accumulator  the  lead  sulfate  is  decomposed 
again  into  Pb  and  PbO2,  as  the  reactions  have  been 
reversed.  In  charging  the  accumulator  the  reactions 
are  at  the  anode 

(64)  PbS04  =  Pb++  +  S04- 

(65)  Pb++  =  Pb  +  2(+) 

Adding  these  two  reactions  together : 
(A)  PbS04  =  Pb  +  2(+)  +  S04— . 
The  reactions  at  the  cathode  are : 

(66)  PbS04  +  2H20  =  2H+  +  SO4—  +  H2Pb02 

(67)  H2PbO2  =  PbO2-  +  2H+ 

(68)  Pb02—  =  Pb02  +  2(-) 

Adding  these  three  reactions  together: 

(E)   PbS04  +  2H20  =  Pb02  +  SO4—  +  4H++  2(-)   ;' 

and  comparing  equation  (A)  with  (5)  we  find  the  two 
negative  charges  liberated  in  (J3)  counterbalanced  by 
the  two  positive  charges  of  (A).  By  the  addition  of 
(A)  to  (B)  it  is  possible  to  represent  the  reactions  occur- 
ring at  the  electrodes  during  charge  and  discharge  of 
the  accumulator: 


CONTROL  OF  REACTIONS 

anode 


charge 


2PbSO4  +  2H2O      <=*       Pb 

discharge 


electrolyte 

+  2so4-  +!H+ 

electrolyte 

Chemical  Affinity.  —  The  preceding  paragraphs  show 
the  influence  of  mechanical,  thermal,  and  electrical  means 
and  the  corresponding  phenomena  have  been  discussed 
as  mass  action,  kinetic  or  free  energy,  and  electro-motive 
force  of  the  atoms.  However  some  reactions  are  affected 
by  another  factor  which  can  not  be  grouped  under  these 
three  headings.  This  fourth  factor  depends  upon  the 
nature  of  the  substance  and  is  expressed  by  a  certain  selec- 
tive tendency  or  chemical  affinity  of  one  type  of  atoms 
for  another  type  of  atoms.  The  chemical  affinity  mani- 
fests itself  in  certain  elements  which  are  grouped  together 
in  the  periodic  chart  (see  appendix).  Thus  the  elements 
located  around  boron  have  an  affinity  for  nitrogen  and 
yield  nitrides  (BN,  etc.),  while  those  in  the  neighbor- 
hood of  lithium  have  an  affinity  for  hydrogen  and  give 
hydrides  (LiH,  etc.).  The  reason  of  chemical  affinity 
is  not  due  to  electro-motive  force  and  is  yet  unknown.1 

Summary.  —  By  physical  means  (mechanical,  thermal, 
electrical)  the  speed  of  a  reaction  can  be  (a)  accelerated, 
(6)  retarded,  (c)  stopped,  or  (d)  reversed. 

Mechanical  means  when  applied  to  ionic  and  molecular 
reactions  control  the  number  of  molecules  per  unit 
volume:  (a)  subdivision  of  solid  substances  (colloids), 
(6)  concentration  of  solutions,  (c)  pressure  of  gases,  (d) 

1  For  a  discussion  of  affinity  see  American  Journal  of  Science,  vol.  46, 
page  490,  1918. 


76  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

catalysers.  In  molecular  reactions  an  increase  in  the 
number  of  molecules  will  increase  the  frequency  of 
molecular  collisions  and  thereby  increase  the  possibility 
of  reaction.  In  ionic  reactions  the  control  of  the  concen- 
tration is  brought  about  by  the  removal  of  anions  or 
cations;  thus  the  possible  formation  of  (a)  water  removes 
H+  and  OH~,  (6)  precipitate  removes  either  cation  or 
anion,  (c)  gas  removes  either  cation  or  anion,  (d)  weak 
acid  removes  H+,  (e)  weak  base  removes  OH~,  (/)  weak 
salt  removes  cation  or  anion,  (g)  complex  ions  remove 
cation  or  anion. 

Thermal  means  are  generally  applied  to  molecular 
reactions  and  consist  in  control  of  temperature  or  the 
velocity  of  molecular  vibrations..  Raising  the  tempera- 
ture increases  the  velocity  of  the  molecules,  thus  increas- 
ing the  frequency  of  molecular  collisions  and  hence 
accelerating  the  speed  of  a  reaction. 

Electrical  means  when  applied  to  ionic  reactions 
control  the  local  separation  of  ions.  The  density  of  the 
electric  current  will  determine  the  speed  of  the  reaction. 

QUESTIONS  AND  PROBLEMS 

1.  What  will  be  the  effect  of  (a)  little,  (b)  much  water  on  the  hydrolysis 
of  BiCl3  (equation  160)  ? 

2.  Write  equations  for  the  reaction  of  bromine  with  (a)  aluminum, 

(b)  iron,  (c)  magnesium,  (d)  sodium,  and  arrange  them  according  to  their 
relative  speeds  as  deduced  from  the  displacement  series. 

3.  By  heating  the  hydroxides  and  carbonates  of  (a)  calcium,  (b)  copper, 

(c)  iron,  (d)  nickel,  (e)  silver,  the  respective  metal  oxides  are  formed. 
Write  equations  and  state  which  compound  requires  the  lowest,  and 
which  compound  the  highest  temperature  for  its  decomposition. 

4.  By  adding  water  to   (a)   Ca3N2,   (b)   Ca3P2,  (c)   Ca3As2,  calcium- 
hydroxide  is  formed.     What  are  the  other  reaction  products  and  which 
of  these  three  reactions  will  proceed  most  rapidly  and  why? 

5.  By  heating  aluminum  with  (a)  antimony,  (b)  arsenic,  (c)  bismuth, 

(d)  phosphorus,  the  respective  binary  compounds  are  formed.     Which 
compound  will  require  the  least  amount  of  heat? 

6.  If  the  metals  Al,  Ga,  Fe,  In,  Mg,  Zn,  are  heated  in  a  current  of 


CONTROL  OF  REACTIONS  77 

chlorine  gas,  the  respective  chlorides  are  formed.  Write  the  equations 
and  state  which  metal  requires  the  lowest,  and  which  metal  requires  the 
highest  temperature  to  start  the  reaction. 

7.  By  heating  TiCl4  with  metallic  Zn  or  Sn,  the  TiCU  is  reduced  to 
TiCl3  and  the  bichlorides  of  Zn  or  Sn  are  formed.     Write  equations 
and  state  which  reaction  occurs  more  readily. 

8.  Arrange  according  to  velocity  the  reactions  of  water  with  metallic 
(a)  calcium,  (b)  copper,  (c)  aluminum,  (d)  lithium,  (e)  iron,  (/)  magnesium, 
(g)  potassium,  (h)  sodium. 

9.  GeO2  can  be  reduced  to  metallic  Ge  with  either  H2  or  C.     Which 
reaction    proceeds   more    readily?     Which    compound    does    hydrogen 
gas  reduce  more  readily  GeO2  or  GeS2?     Write  equations  for  each. 

10.  Predict  what  will  happen  if  metallic  magnesium  is  heated  with 
(a)  SiO2,  (b)  GeO2,  (c)  PbO2.     Which  reaction  proceeds  more  readily? 

11.  Carbon  dioxide  can  be  reduced  to  carbon  by   (a)   calcium,    (6) 
magnesium,    (c)    potassium,    (d)    sodium.     Write    the    equations    and 
arrange  them  in  the  order  of  their  speed  of  reaction. 

12.  Point  out  which  metals  are  soluble  in  sulfuric  acid  under  the  evolu- 
tion of  hydrogen  gas:  (a)  aluminum,  (b)  copper,  (c)  iron,  (d)  magnesium, 
(e)  silver,  (/)  tin,  (g)  zinc? 

13.  Which  element  reduces  HAuCl4  most  readily :  (a)  copper,  (6)  iron, 
(c)  phosphorus,  (d)  tin? 

14.  Arrange  the  following  oxides  in  the  order  in  which  carbon  will 
reduce  them  most  readily  to  metals:  (a)  aluminum  oxide,  (6)  iron  oxide, 
(c)  lead  oxide,  (d)  magnesium  oxide,  (e)  tin  oxide,  (/)  zinc  oxide.     Write 
the  respective  equations. 

16.  Give  a  reason  why  iodine  replaces  chlorine  in  the  equation: 
2KC1O3  +  I2  =  2KIOs  -f  C12.  (This  reaction  occurs  in  the  presence 
of  HNOs  as  catalyser). 

16.  By  heating  metallic  Mg  and  RbOH  the  following  reaction  occurs: 
2RbOH  +  2Mg  =  2Kb  +  2MgO  +  H2.     Why  is  hydrogen  gas  formed? 
Why  is  it  not  possible  for  the  reaction  Mg(OH)2  -f-  2Kb  =  2RbOH  -f 
Mg  to  occur? 

17.  If  ammonium  bichromate  is  heated  green  Cr2Oa  is  formed.     What 
are  the  possible  reaction  products  ?    How  could  you  test  by  experiment 
whether  the  oxygen  or  nitrogen  is  oxidized  ?     (Write  two  or  more  equa- 
tions of  possible  reactions.) 


CHAPTER  VI 
TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS 

Reactions  and  their  equations  are  divided  into  types. 
Thus,  if  a  complex  molecule  is  broken  apart  into  simpler 
or  elementary  constituents,  the  process  is  called  analysis; 
the  reverse  process  of  constructing  or  building  up 
complex  molecules  from  elementary  constituents  is 
called  synthesis;  while  the  exchange  of  parts  of  one 
molecule  with  parts  of  another  molecule  is  termed 
metathesis.  The  first  two  terms  are  explained  by  the 
following  scheme: 


Type  of 
reaction 


ANALYSIS 


marble 

(calcium  carbonate) 
CaCO3 


SYNTHESIS 


lime 

(calcium  oxide) 
CaO 


y       \ 

calcium    oxygen 
Ca  O 


calcium  oxide 
CaO 


\ 


carbon- 
dioxide 
C02 


/  \ 

carbon    oxygen 
C  O2 


X 

carbon  dioxide 
C02 


calcium  carbonate 
CaC03 

78 


Type  of 
substance 

complex, 
compound 


simple 
compound 


element 


simple 
compound 


complex 
compound 


TYPES  OP  REACTIONS  AND  THEIR  EQUATIONS  79 

Analysis  in  the  above  example  involves  the  reactions: 

(69)  CaCO3->CaO  +  CO2 

(70)  2CaO-»2Ca  +  O2 

(71)  CO2-+C  +  O2 

while  synthesis  is  expressed  by  the  equations: 

(72)  CaO  +  CO2-»CaCO3 

(73)  2Ca  +  O2-»2CaO 

(74)  C  +  O2 


An  examination  of  these  two  sets  of  equations  reveals 
the  fact  that  in  (69)  and  (72)  there  are  no  changes  in  the 
valence  number  of  the  elements  involved,  while  in  (70), 
(71),  (73),  and  (74)  there  is  a  change  in  the  valence 
number  of  the  elements,  therefore  analysis  and  synthesis 
may  or  may  not  involve  oxydation  and  reduction.  It 
is  also  evident  that  (72)  is  the  reverse  of  (69),  (73)  the 
reverse  of  (70),  (74)  the  reverse  of  (71).  An  analytical 
reaction  of  type  (69)  is  termed  decomposition,  while  the 
reverse  synthetical  reaction  (72)  is  addition  —  in  these 
two  types  there  is  no  change  of  valency.  The  analytical 
reaction  of  type  (70)  or  (71)  is  known  as  reduction,  while 
the  reverse  synthetical  reaction  (73)  or  (74)  is  usually 
termed  oxydation.  But  the  terms  oxydation  and  reduc- 
tion are  ambigiuous  for  they  also  mean  an  increase  or 
decrease  in  the  valence  number  of  an  element,  therefore 
it  is  more  exact  to  use  combination  for  oxydation,  and 
division  for  reduction  when  speaking  of  reactions. 

An  example  of  metathesis: 

(75)         CaCl2  +  Na2CO3  =  CaCO3  +  2NaCl 

shows  that  there  is  an  exchange  of  parts  in  a  molecule 
without  change  in  the  valence  number.  One  of  the 
most  common  types  of  metathesical  reactions  is  neutrali- 
zation which  has  previously  been  defined  as  the  reaction 


80  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

between  an  acid  and  a  base  to  give  water  and  a  salt. 
An  example  is: 

(76)  Ca(OH)2  +  2HC1  =  CaCl2  +  2H2O 
while  the  reverse  reaction  is  hydrolysis : 

(77)  CaCl2  +  2H2O  =  Ca(OH)2  +  2HC1 

The  interrelation  of  these  different  types  of  reactions 
is  shown  in  the  following  diagram  in  which  the  arrow 
indicates  the  direction  of  the  reaction,  M  and  X  a  metal 
or  positive  radical,  N  and  Y  a  non-metal  or  negative 
radical. 

TYPES  OF  REACTIONS 

A.  Reactions  involving  no  oxidation  and  reduction 
(i.e.  no  increase  or  decrease  in  the  valence  numbers  of 
the  elements): 

I.  ADDITION 

MN  +  XN  =  MN.XN 


II.  DECOMPOSITION 


=  MXN2 


III.  METATHESIS  MN  +  XY  =  MY  +  XN 


IV.  NEUTRALIZATION 
V.  HYDROLYSIS  MOH+HN-MN+HOH 


B.  Reactions  involving  oxidation  and  reduction  (i.e. 
increase  or_  decrease  in  the  valence  numbers  of  the 
elements) : 

VI.  COMBINATION 


VII.  DIVISION  M  MN 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  81 

VIII.  DISPLACEMENT       |  M  +  XN  =  MN  +  X 

?=  -•=*       \  N  +  MY  =  MN  +  Y 

IX.  SUBSTITUTION     }    |2M  +.XN  =  MN  +MX 


X.  RESTITUTION 


2N  +  MY  =  MN  +  YN 


Examples  of  these  ten  types  of  reactions  are  given  in 
the  text  and  later  the  complex  reactions  are  discussed. 
Complex  reactions  are  those  which  involve  two  or  more 
types  of  reactions. 

I.  ADDITION 

Ordinarily  two  binary  compounds  combine  and  form 
a  more  complex  compound.  The  general  type  is  MX  + 

NX  =  MX.NX  =  MNX2.     The  oxysalts  of  metals  are 
addition  products: 

(78)  K2O  +  SO3  =  (K2O.SO3)  =  K2SO4 

(79)  CaO  +  C02  =  (CaO.CO2)  =  CaCO3 

Many  sulfides  form  compounds  of  this  type: 

(80)  PbS  +  Sb2S3  =  (PbS.Sb2S3)  =  PbSb2S4 

(81)  3CuS  +  Sb2S3  =  (3CuS.Sb2S3)  =  2Cu3SbS3 

Some  acids  and  bases  may  be  considered  addition  pro- 
ducts of  water  and  a  metallic  or  non-metallic  oxide: 

(82)  H20  +  N206  =  (H2O.N2O6)  =  2HNO3 

(83)  H2O  +  SO3  =  (H2O.SO3)  =  H2SO4 

(84)  H2O  +  CaO  =  (H2O.CaO)  =  Ca(OH)2 

(85)  H2O  +  Na2O  =  (H2O.Na2O)  =  2NaOH 

(86)  H2O  +  NH3  =  (H2O.NH3)  =  NH4OH 

Other  compounds  belonging  to  addition  products  are: 

(87)  3PC16  +  P2O5  =  5POC13' 

6 


82  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

(88)  NH3  +  HC1  =  NH4C1 

(89)  Ni++  +  6NH3  =  Ni(NH3)6++ 

(90)  Cu++  +  4NH4OH  =  Cu(NH3)4++  +  4H2O 

(91)  AgCl  +  2NH4OH  =  2Ag(NH3)2Cl  +  2H2O 

(92)  Zn(OH)2  +  2NH4C1  +  4NH3  =  Zn(NH3)8Cl2  + 

2H2O 

(93)  NaOH  +  CO  =  NaHC02  or  HCOONa 

(94)  As2S3  +  3(NH4)2S  =  2(NH4)3AsS3 

(95)  SnS  +  (NH4)2S2  =  (NH4)2SnS3 

(96)  SnS2  +  (NH4)2S  =  (NH4)2SnS3 

(97)  Fe(CN)2  +  4KCN  =  K4Fe(CN)6 

(98)  Zn(CN)2  +  2KCN  =  K2Zn(CN)4 

(99)  Ni(CN)2  +  2KCN  =  K2Ni(CN)4 

(100)  Co(N02)3  +  3KNO2  =  K3Co(NO2)6 

II.  DECOMPOSITION 

Decomposition  is  a  chemical  reaction  in  which  a 
complex  compound  is  broken  apart  into  simpler  com- 
pounds without  any  change  in  the  valence  numbers  of 
the  elements  involved — it  is  the  reverse  reaction  of 
addition : 

(101)  NH4C1  =  NH3  +  HC1 

(102)  Cu(OH)2  =  CuO  +  H2O 

(103)  H4SnO4  =  H2SnO3  +  H2O 

(104)  Cu(NO3)2  =  CuO  +  N205 

(105)  Te(OH)6  =  TeO3  +  3H2O 

(106)  2B(NH2)3  =  B2(NH)3  +  NH3  =  2BN  +  4NH3 

To  this  type  belongs  dehydration,  which  is  the  reaction 
taking  place  when  crystals  containing  water  of  crystal- 
lization are  heated  and  lose  part  or  all  of  their  water. 
Dehydration  usually  proceeds  in  steps,  depending  upon 
temperature,  thus 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  83 

beginning  at  19°C.: 

MgSO4.7H2O  =  MgSO4.6H^O  +    H2O 
beginning  at  38°C.: 

MgSO4.6H2O  =  MgSO4.2H2O  +  4H2O 
beginning  at  112°C.r 

MgSO4.2H2O  =  MgSO4.H2O   -f    H2O  * 
beginning  at  203°C.: 

MgSO4.H2O  =  MgSO4  +    H2O 

The  complete  dehydration  is  expressed  by: 
(107a)          MgSO4.7H2O  =  MgSO4  +  7H2O 

In  this  case  the  dehydrated  magnesium  sulfate,  when 
heated  to  about  400°C.,  will  begin  to  decompose  accord- 
ing to 

(107b)  MgSO4  -  MgO  +  SO3 

Many  other  sulfates  behave  similarly.  The  dehydration 
and  decomposition  of  copper  sulfate  is  expressed  in  the 
equations : 

CuSO4.5H2O  =  CuSO4.3H2O  -f  2H2O,  begins  at  27°C.  \ 

CuSO4.3H2O  =  CuSO4.H2O    +  2H2O,  begins   at  93°C.  |  dehydration 

CuSO4.H2O    =  CuSO4  +  H2O,    begins  at  11 5°C.  J 

2CuSO4  =  CuSO4.CuO    +  SO3,     begins  at  660°C.  |  decomposi- 

CuSO4.CuO    =  2CuO  4-  SO3,  begins  at  710°C.  J         tion 

The  complete  dehydration  is  thus: 
(108a)  CuSO4.7JI2O  =  CuSO4  +  7H2O 

and  the  complete  decomposition: 
(108b)  CuSO4  =  CuO  +  SO3 

UI.  METATHESIS 

Metathesis  is  a  common  reaction.     The  general  type 
MN  +  XY  =  MY  +  XN  indicates  a  double  decomposi- 


84  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

tion  in  which  an  exchange  of  the  elements  or  radicals 
has  taken  place. 
Some  examples  are: 

(109)  AgNO3  +  NaCl  =  AgCl  +  NaNO3 

(110)  CaO  +  2HC1  =  CaCl2  +  H2O 

(111)  CaS  +  2HC1  =  CaCl2  +  H2S 

(112)  Ca3P2  +  6HC1  =  CaCl2  +  2PH3 

(113)  A12O3  +  6HC1  =  2A1C13  +  3H2O 

(114)  A12S3  +  6HC1  =  2A1C13  +  3H2S 

(115)  Al2Se3  +  6HC1  =  2A1C13  +  3H2Se 

(116)  Al2Te3  +  6HC1  =  2A1C13  +  3H2Te 

(117)  FeS  +  2HC1  =  FeCl2  +  H2S 

(118)  2Na3As  +  6HC1  =  2AsH3  +  6NaCl 

(119)  SnCl2  +  H2S  =  SnS  +  2HC1 

(120)  CaO  +  H2S  =  CaS  +  H2O 

(121)  CaCO3  +  Na2S  =  CaS  +  Na2CO3 

(122)  2KC1  +  H,SiF,  =  K2SiF6  +  2HC1 

(123)  3CaO  +  Na3AlF6  =  3CaF2  +  Na3AlO3 

(=  Al(ONa),) 

(124)  CaSiO3  +  3H2F2  =  SiF4  +  CaF2  +  3H2O 

(125)  WO3  +  2KOH  =  K2WO4  +  H2O 

(126)  K2WO4  +  H2SO4  +  H2O  =  K2SO4  +  WO(OH), 

(127)  2A1(OH)3  +  3CS2  =  A12S3  +  3H2S  +  3CO2 

(128)  H2SO4  +  P2OB  =  SO3  +  2HPO3 

(129)  SnCl2  +  Na2CO3  +  H2O  =  Sn(OH)2  + 

2NaCl  +  CO2 

(130)  CaC2  •+  2H20  =  C2H2  +  Ca(OH)2 

(131)  A14C3  +  12H2O  =  3CH4  +  4A1(OH)3      • 

(132)  PH4I  +  KOH  =  PH3  +  KI  +  H2O 

(133)  2H3BO3  +  3PC15  =  B2O3  +  3POC13  +  6HC1 

(134)  2CuCl  +  Na2CO3  =  Cu2O  +  2NaCl  +  CO2 

(135)  CuClo  +  Na2CO3  =  CuCO3  +  2NaCl 

(136)  Mo03  +  2KOH  =  K2MoO4  +  H2O 

(137)  Au,O,  +  2KOH  =  2KAu02  +  H20 


TYPES  OF  EEACTIONS  AND  THEIR  EQUATIONS  85 

In  the  preceding  list  only  molecular  equations  are 
given — some  of  which  can  be  put  into  ionic  form.  Typi- 
cal ionic  equations  of  metathesic  reactions  are : 

(138)  Ag+  +  Cl-  =  AgCl 

(139)  Ba++  +  SO4—  =  BaS04 

(140)  Ba++  +  CrO4—  =  BaCrO4 

(141)  Cu++  +  20H-  =  Cu(OH)2 

(142)  Co++  +  2OH-  =  Co(OH)2 

(143)  Zn++  +  2OH-  =  Zn(OH)2 

(144)  Zn(OH)2  +  2OH-  =  Zn02-  +  2H2O 

(145)  Fe++  +  2CN-  =  Fe(CN)2 

(146)  Ni++  +  2CN-  =  Ni(CN)2 . 

(147)  Zn++  +  2CN-  =  Zn(CN)2 

(148)  2Cu++  +  Fe(CN)6 =  Cu2Fe(CN)6 

(149)  Cu(OH)2  +  4NH4OH  =  Cu(NH3)4++  +  20H- 

+  4H20 

(150)  2Na3SbS  +  6H+  =  Sb2S6  +  6Na+  +  3H2S 

(151)  BaO2  +  2H+  =  Ba++  +  H2O2 

(152)  PtCl6-  +  2K+  =  K2PtCl6 

(153)  Cu++  +  H2S  =  CuS  +  2H+ 

IV.  NEUTRALIZATION 

As  previously  stated,  the  general  equation  for  neu- 
tralization is  H+  +  OH-  =  H2O.  This  equation  is 
irrespective  of  the  acid  and  base  employed.  The  salt 
formed  differs  in  accordance  with  the  acids  and  bases 
used,  but  the  formation  of  water  is  the  principal  earmark 
of  neutralization  (compare  equations  4  to  15).  By 
attaching  a  negative  nonmetal  or  acid  radical,  N,  to  the 
positive  hydrogen  ion  and  to  the  negative  hydroxyl  ion 
a  positive  metal  or  basic  radical,  M,  the  general  type  of 
neutralization  is  M(OH)  +  HN  =  MN  +  HOH  and 
hence  it  is  a  metathesical  reaction.  By  substituting  any 


86  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

metal  or  positive  radical  for  M,  and  any  nonmetal  or 
negative  radical  for  N,  a  great  number  of  particular 
reactions  are  possible,  some  of  which  are  given  in  equa- 
tions 4  to  15. 

V.  HYDROLYSIS 

Hydrolysis  is  a  metathesical  change  brought  about 
by  the  action  of  water  and  is  the  reverse  reaction  of 
neutralization, — the  general  type  is  MN  +  HOH  = 
M(OH)  +  HN.  Hydrolysis  takes  place  when  the  salt 
of  a  weak  acid,  of  a  weak  base,  or  a  compound,  whose 
constituents  are  not  far  apart  in  the  displacement  series, 
is  dissolved  or  brought  into  contact  with  water. 

Hydrolysis  of  salts  of  a  weak  acid : 

(154)  CH3COONa  +  H2O  =  NaOH  +  CH3COOH 

(155)  Na2CO3  +  H2O  =  NaOH  +  NaHCO3 

(156)  NaHCO3  +  H2O  =  NaOH  +  H2CO3  =  NaOH  + 

H2O  +  CO2 

(157)  2BaS  +  2H2O  =  Ba(OH)2  +  Ba(SH)2 

(158)  Ba(SH)2  +  2H2O  =  Ba(OH)2  +  2H2S 

Hydrolysis  of  salts  of  a  weak  base: 

(159)  CuSO4  +  2H2O  =  Cu(OH)2  +  H2SO4 

(160)  BiCl3  +  H2O  =  BiOCl  +  2HC1 

(161)  BiOCl  +  2H2O  -  Bi(OH)3  +  HC1 

(162)  Aid,  +  3H2O  -  A1(OH)3  +  3HC1 

Hydrolysis  of  compounds  whose  constituent  elements 
are  not  far  apart  in  the  displacement  series : 

(163)  PC13  +  3H20  =  3HC1  +  P(OH)3  (  =  H3PO3) 

(164)  PC15  +  5H20  =  5HC1  +  P(OH)5  (  =  H,PO4  + 

H20). 

(165)  SbCl3  +  H20  =  SbOCl  +  HC1 

(166)  SiCl4  +  2H2O  =  SiO2  +  2HC1 

(167)  CC14  +  H20  =  COC12  +  2HC1 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  87 

(168)  BN  +  2H2O  =  HBO2  +  NH3 

(169)  Ca3As2  +  6H2O  =  2AsH3  +  3Ca(OH)2 

(170)  WC16  +  5H2O  =  H4WO5  +  6HC1 

(171)  PB3  +  3H2O  =  3HBr  +  P(OH)3 

(172)  PBr5  +  H2O  =  POBr3  +  2HBr 

(173)  POC13  +  3H2O  =  H3PO4  +  3HC1 

Under  certain  conditions  even  distinct  polar  compounds 
hydrolyse: 

(174)  MgCl2  +  H2O  =  MgO  +  2HC1 

(175)  2NaCl  +  H2O  =  Na2O  +  2HC1 

These  polar  compounds  are  called  "strong"  salts  while 
the  compounds  enumerated  in  the  list  which  have  con- 
stituents near  together  in  the  displacement  series,  are 
termed  "weak"  salts.  The  former  ionize  strongly, 
the  latter  ionize  but  slightly  if  at  all. 

VI.  COMBINATION 

Combination  is  the  union  of  two  elements  resulting 
in  the  formation  of  a  compound:  M  +  N  =  MN. 
While  this  equation  apparently  is  the  simplest  type  of 
reaction,  it  involves  the  principles  of  oxidation  and 
reduction  for  M  is  usually  oxidized  and  N  is  reduced. 
Ordinarily  combination  is  called  oxidation,  and  the 
reverse  reaction  is  termed  reduction.  In  a  strict  sense, 
however,  combination  as  well  as  division  involves 
oxidation  and  reduction.  The  common  terminology, 
while  practical  in  some  respects,  is  thus  ambigious  and 
should  be  avoided.  Some  examples  of  combination  or 
" oxidation"  are: 

(176)  2H2  +  O2  =  2H2O 

(177)  Ca  +  C12  =  CaCl2 

(178)  Fe  +  S  =  FeS 


.:-:  = 


(202)  4KMnO<  =  2K^IM>,  +  3O* 

203  Mn(OH),  =  HMnO.  +  H,O 

204  2Pb(XO,),  =  2PbO 


A  related  reaction  is  dinociaiion,  that 

apart    of    molecules    at 


pbee  (see  Chapter  5).    ExamtpU:  At 

(206)  2Fe-O,  =  4Te  +  3Os 

At  low  temperature: 

(206)  4Fe  +  3O-  =  2Fc 


Displacement  is  a  A*«Mg-J  reaction  in 

imimatlMar  element  in  a  compoiincL    The 
for  metals  is  M  +  YX  =  MX  +  T; 

N   -  YX  =  YX  -  X      E:---  .1  -vil 
can  replace  another  metal,  or  a  nonmetal  can  replace 

metal,  M,  is  more  electro-positive  than  Ike  metal,  Y. 


nonmetalrX.    The  electro-positive  or  negative  < 
appendix).    Kramples  off  this  type   of  reactinag  are: 


(.207)  Cu  +  HgCls  =  OiCI,  +  Hg 

(208)  Fe  +  CnCls  -  Fed,  +  ». 

(209)  Zn  +  Fed,  =  ZnO,  +  Fe 

(210)  2K  +  ZnCl,  =  2KQ  +  & 

211  Zn -I- 2AgCl  =  ZnCl.  +  Ag 

212  Fe  +  PbS  -  FeS  +  Pb 

212a  +  K«Fe(C30.  -  «KCN  +  Fe 


90  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

Displacement  by  nonmetals: 

(213)  Br2  +  2AgI  =  2AgBr  +  21 

(214)  C12  +  2AgBr  =  2AgCl  +  Br2 

(215)  F2  +  2AgCl  =  2AgF  +  C12 

(216)  3F2  +  3H2O  =  6HF  +  O3 

(217)  2F2  +  2H2O  =  2H2F2  +  02 

(218)  O2  +  4HC1  =  2H2O  +  2C12 

(219)  O2  +  2MgCl2  =  2MgO  +  2C12 

(220)  5O2  +  4PBr3  =  2P2O5  +  6Br2 

(221)  Br2  +  H2S  =  2HBr  +  S 

The  many  displacement  reactions  which  involve  hy- 
drogen can  be  grouped  into  two  types.  In  the  first  type 
of  displacement  the  hydrogen  of  acids  or  bases  is  being 
replaced  by  a  more  electro-positive  metal.  In  this  case 
hydrogen  is  reduced  and  hence  acts  as  an  oxidizing  agent : 

(222)  Zn  +  H2S04  =  ZnSO4  +  H2 

(223)  2A1  +  6HC1  =  2A1C13  +  3H2 

(224)  Mg  +  2HNO3  =  Mg(NO3)2  +  H2 

(225)  Zn  +  2KOH  =  Zn(OK)2  +  H2  (=  K2ZnO2) 

(226)  Al  +  2KOH  =  KA1O2  +  K  +  H2 

(227)  2Mg  +  2KOH  =  MgO  +  2K  +  H2 

In  the  other  cases  the  hydrogen  is  oxidized,  and  so  acts 
as  a  reducing  agent: 

(228)  2Na2O  +  H2  =  2NaOH  +  2Na 

(229)  CuO  +  H2  =  H2O  +  Cu 

(230)  Fe2O3  +  3H2  =  3H20  +  2Fe 

(231)  2AgCl  +  H2  =  2HC1  +  2Ag 

(232)  W03  +  3H2  =  3H20  +  W 

The  first  set  of  reactions  (222-227),  in  which  hydrogen 
is  replaced  by  a  metal,  depends  upon  the  electromotive 
force  of  the  metal  and  the  following  general  rule  is 
deduced:  Whenever  the  metal,  M,  is  more  positive 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS       91 

than  hydrogen  (see  displacement  series)  it  will  take 
the  charge  from  the  hydrogen  ion  and  liberate  hydrogen 
gas,  hence  these  metals  are  all  soluble  in  acids  according 
to  the  general  equation  2M  +  2H+  =  2M+  +  H2. 

IX.  SUBSTITUTION 

Substitution  is  a  chemical  reaction  which  resembles 
displacement;  however,  the  element  which  interacts 
with  a  compound  does  not  liberate  the  replaced  element 
but  combines  with  it  according  to  the  general  type 
2X  +  MN  =  XN  +  XM  in  the  case  of  metals,  and 
2Y  +  MN  =  MY  +  NY  in  the  case  of  nonmetals : 

(233)  2NH3  +  HgCl2  =  NH4C1  +  HgNH2Cl 

(234)  C12  +  CH4  =  HC1  +  CH3C1 

(235)  C12  +  H2O  =  HC1  +  HOC1 

(236)  61  +  5AgF  =  IF5  +  5AgI 

(237)  302  +  2PbS  =  2S02  +  2PbO 

A  related  reaction  is: 

(238)  4Na  +  3SiF4  =  Si  +  2Na2SiF6 

X.  RESTITUTION 

The  reverse  reaction  of  substitution  is  re-substitution 
or  restitution,  the  general  type  being  MN  +  MY  = 
M  +  NY.  To  this  class  belong: 

(239)  2PbO  +  PbS  =  3Pb  +  SO2 

(240)  2CuO  +  Cu2S  =  4Cu  +  SO2 

(241)  Cu2S  +  CuSO4  =  3Cu  +  2SO2 

(242)  CO  +  H20  =  H2  +  C02 

Similar  and  related  reactions  are: 

(243)  3BaSO4  +  BaS  =  4BaO  +  SO2 

(244)  BaCO3  +  C  =  BaO  +  2CO 
(246)    CaC2  +  N2  =  CaCN2  +  C 


92  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

(246)  4BaSO4  +  4C  =  4BaO  +  S02  +  4CO 

(247)  BaCO3  +  3Mg  +  C  =  BaC2  +  3MgO 

(248)  2KNO3  +  10K  =  6K2O  +  N2 

XI.  OXIDATION  AND  REDUCTION 

There  are  many  reactions  not  classified  under  the 
previous  titles  which  are  characterized  by  oxidation 
and  reduction.  These  reactions  must  be  studied  and 
in  each  case  it  should  be  determined  which  elements 
are  oxidized  and  which  are  reduced,  for  only  thus  is  it 
possible  to  get  a  clear  conception  of  the  meaning  of 
oxidation  and  reduction. 

A  number  of  such  reactions  are : 

(249)  2SO2  +  02  =  2SO3 

(250)  2KNO3  =  2KNO2  +  O2 

(251)  PC16  =  PC13  +  Cl, 

(252)  2CrO3  +  2NH3  =  Cr203  +  N2  +  3H2O 

(253)  2CrO3  +  12HC1  =  2CrCl3  +  3C12  +  6H2O 

(254)  2CuSO4  +  4KCN  =  2CuCN  +  2K2SO4  +  C2N2 

(255)  4Co(OH)3  +  4H2SO4  =  4CoSO4  +  O2  +  10H2O 

(256)  3KC1O  =  KC1O3  +  2KC1 

(257)  2KC1O3  =  KC1O4  +  KC1  +  O2 

(258)  Na2SO4  +  40  =  Na2S  +  4CO 

(259)  2MnO2  +  2K2O  +  O2  =  2K2MnO4 

(260)  H2TeCl6  +  2SO2  +  4H2O  =  Te  +  2H2SO4  + 

6HC1 

(261)  2KOH  +  2NO2  =  KNO2  +  KNO3  +  H2O 

(262)  2KOH  +  2C1O2  =  KC1O3  +  KC102  +  H2O 

(263)  2P  +  5Br2  +  6H2O  =  lOHBr  +  2HPO3 

(264)  Na2S2O3  +  Br2  +  H2O  =  Na2SO4  +  2HBr 

(265)  10NO  +  6KMnO4  +  9H2SO4  =  10HNO3  + 

6MnSO4  +  3K2SO4  +  4H2O 

(266)  Au  +  4HC1  +  HNO,  =  HAuCl4  +  NO  +  2H20 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  93 

(267)  PbO2  +  HNO3  +  HNO2  =  Pb(N03)2  +  H2O 

(268)  4Hg2CrO4  =  2Cr2O3  +  4Hg2O  +  3O2 

(269)  4Hg2O  =  8Hg  +  2O2 

(270)  2HgCl2  +  SO2  +  2H2O  =  2HC1  +  2HgCl 

+  H2S04 

(271)  P4  +  3KOH  +  3H2O  =  PH3  +  3KH2PO2 

(272)  Ag3As  +  3AgNO3  +  3H2O  =  6Ag  +  As(OH)3 

+  3HNO3 

(273)  3Ca(PO3)2  +  IOC  =  P4  +  Cas(P04)2  +  10CO 

(274)  2Ca(P03)2  +  2Si02  +  IOC  =  2CaSiO3  + 

10CO  +  P4 

(275)  2Ca3(PO4)2  +  6SiO2  +  IOC  =  6CaSiO3  + 

10CO  +  P4 

(276)  TiO2  +  2C  +  3C12  =  TiCl4  +  2COC1 

(277)  4RbO2  +  4H2  =  4RbOH  +  2H2O  +  O2 

(278)  2RbO2  +  H2O  =  2RbOH  +  O2 

(279)  4K2FeO4  +  10H2O  =  4Fe(OH)3  +  3O2  +  8KOH 

(280)  2K4Fe(CN)6  +  C12  =  2K3Fe(CN)6  +  2KC1. 

(281)  K2Cr207  +  3H2  =  Cr2O3  +  K2O  +  3H2O 

(282)  NC13  +  3H20  =  NH3  +  3HOC1 

(283)  2S2C12  +  2H2O  =  SO2  +  3S  +  4HC1 

(284)  SC14  +  2H2O  =  SO2  +  4HC1 

(285)  2Co(OH)2  +  Br2  +  2NaOH  =  2Co(OH)3  + 

2NaBr 

(286)  2H2S203  +  21  =  2HI  +  H2S4O6 

(287)  5N2H4  +  4KIO3  +  4HC1  =  5N2  +  4KC1  + 

41  +  12H2O 

(288)  4HNO3  +  4CH2O  =  4NO2  +  3H2O  +  CO2 

(289)  NH4N03  =  N2O  +  2H2O 

(290)  H2N2O2  =  N2O  +  H2O 

(291)  HNO2  +  NH2OH  =  H2N2O2  +  H2O 

(292)  T1C1  +  Cl,  =  T1C13 

(293)  2Mn(OH)3  +  MnO2  +  2H2  =  3MnO  +  5H2O 

(294)  2(NH4)2MoO3  +  O2  =  2MoO3  +  4NH3  +  2H2O 


94  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

The  above  reactions  are  all  in  the  non-ionic  or  molecu- 
lar form;  it  is  a  helpful  exercise  to  transcribe  these  equa- 
tions into  the  ionic  form  in  cases  where  the  substances 
are  ionized.  ^Examples  of  typical  ionic  equations  are: 
Oxidation  by  H2O2: 


(296)  ^Mn++  +  H2O2  =  ^OH~  =  $Mn(OH)3 

(296)  2Fe++  +  H202  +  40H~  =  2Fe(OH)3 

(297)  2Co++  +  H2O»  +  4OH-  =  2Co(OH)3 

(298)  2Cr++  +  H2O2  +  4QH-  =  2Cr(OH)3 

(299)  ^KI  +  H202  =  2KOH  +  I2 

The  metals  Cr  and  Mn  can  be  further  oxidized: 

(300)  2Mn(OH)3  +  H202  =  2Mn02  +  4H2O 

(301)  2MnO2  +  H2O2  +  2OH~  =  2Mn03-  +  2H2O 

(302)  2Cr(OH)3  +  20H-  =  2Cr02-  +  4H2O 

(303)  2CrO2-  +  3H2O2  +  *OH-  =  2CrO4-  +  4H2O 

By  adding  the  equations  for  manganese  (295,  300)  and 
chromium  (298,  302,  303)  together,  the  total  equation 
of  the  reaction  taking  place  becomes: 


(304)  2Mn++  +  2H2O2  +  4OH-  =  2Mn02  +  fH2O 

(305)  2Cr++  +  4H202  +  80H~  =  2CrO4-  +  8H2O 

As  shown  by  these  equations,  hydrogen  peroxide  will  act 
as  an  oxidizing  agent  in  alkaline  solution.  In  acid 
solution,  however,  it  will  act  as  a  reducing  agent: 

(306)  Cr2O7—  +  3H2O2  +  8H+  =  2Cr+++  +  7H2O  + 

802 

(307)  2MnO4-  +  3H202  +  ?H+  =  2Mn02  +  4H2O  +  30 

(308)  2MnO2  +  2H202  +  4H+  =  2Mn++  +  4H20  + 

202 

By  adding  the  last  two  equations  (307,  308)  together, 
and  cancelling  Mn02  which  appears  on  both  sides,  the 
total  equation  becomes: 


2 


.    TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  95 

j 

(309)  2MnO4-  +  5H2O2  +  6H+  =  2Mn++  +  8H2O  + 

5O2 
Reduction  by  NO2~: 

(310)  Cr,OT—  +  3NO2-  +  8H+  =  2Cr+++  +  3NO3- 

+  4H2O 

(311)  2MnO4~  +  5NO2~  +  6H+  =  2Mn++  +  5NO3~ 

+  te2o 

Reduction  by  HI : 

(312)  Cr2O7—  +  6HI  +  8H+  =  2Cr+++  +  3I2  +  7H2O 

(313)  2MnO4-  +  10HI  +  6H+  =  2Mn++  +  5I2  +  SH2O 

(314)  H2SO4  +  2HI  =  H2SO3  +  I»  +  H20 

(315)  H2SO3  +  tel  =  H2S  +  $1,  +  $H2O 

(316)  H2SO4  +  SHI  =  H2S  +  4I2  +  4H2O 

Reduction  by  Fe++: 

(317)  Cr2O7—  +  6Fe++  +  14H+  =  2Cr+++  +  6Fe+++ 

+  7H20 

(318)  MnO4-  +  5Fe++  +  8H+  =  Mn++  +  5Fe+++  + 

4H20 

(319)  3Hg++  +  6Fe++  =  3Hg  +  6Fe+++ 

Reduction  by  H2S: 

(320)  H3AsO4  +  H2S  =  H3AsO3  +  S  +  H2O 

(321)  2H3AsO3  +  3H2S  =  As2S3  +  fH2O 

(322)  2H3AsO4  +  |H2S  =  As2S3  +  ?S  +  te2O 

Oxidation  by  HN03: 

(323)  3Zn  +  8H+  +  2NO,-  =  3Zn++  +  2ND  +  4H2O 

(324)  3Cu  +  8H+  +  2NO3-  =  3Cu++  +  2NO  +  4H2O 

(325)  3Hg  +  8H+  +  2NO3~  =  3Hg++  +  2NO  +  4H2O 

(326)  3Sn  +  4H+  +  4NO3-  =  3H2SnO3  +  4NO 

(327)  3PbS  +  8H+  +  2NO3~  =  3Pb++  +  3S  +  2NO 

+  4H20 


96  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

Oxidation  by  H2S04: 

(328)  Cu  +  H2S04  =  CuO  +  SO2  +  H2O 

(329)  CuO  +  2H+  =  Cu++  +  H20 

(330)  Cu  +  4H+  +  SO4—  =  Cu++  +  SO*  +  2H2O 

Oxidation  by  HNO2: 

(331)  Co(N02)2  +  2HN02  =  Co(NO2)3  +  NO  +  H2O 
Oxidation  and  reduction  of  Hg~: 

(332)  2Hg+  +  H2S  =  Hg  +  HgS  +  2H+ 

(333)  2HgCl  +  2NH3  =  NH4C1  +  HgCNH2)Cl  +  Hg 

(334)  2Hg+  +  2CN-  =  Hg  +  Hg(CN)2 

The  student  should  devise  laboratory  experiments  to 
prove  the  correctness  of  these  reactions,  and  should  be 
able  to  predict  what  will  happen  if  certain  chemicals 
react  upon  each  other.  He  should  tell  whether  the 
solutions  must  be  acidified  or  made  alkaline;  he  should 
know  what  color-changes  or  precipitations  to  expect,  and 
be  able  to  write  the  specific  non-ionic  equations  for  the 
above  ionic  reactions. 

XII.  COMPLEX  REACTIONS 

There  are  many  cases  in  which  the  reaction  products 
interact  upon  each  other  and  produce  a  second  set  of 
reaction  products.  The  equations  of  these  reactions 
seem  most  complex  when  secondary  reactions  are 
included  in  a  single  equation;  but  this  complexity  disap- 
pears when  more  than  one  equation  is  written.  The 
following  shows  the  various  steps  of  such  reactions,  in 
different  equations,  and  these,  added  together,  form  the 
equation  of  a  complex  reaction. 

(335)  2A1C13  +  3H20  =  A1(OH)3  +  6HC1  (hydrolysis) 

(336)  3Na2S203  +  6HC1  =  GNaCl  +  3S  +  3SO2 

(decomposition) 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS  97 

Added  together  and  cancelling  HC1  leaves: 

(337)  2A1C13  +  3Na2S203  +  3H2O  =  Al(OH),  + 

6NaCl  +  3S  +  3SO2 

Reduced  to  the  ionic  form  337  becomes : 

(338)  2A1+++  +  3S2O3—  +  3H2O  =  Al(OH),  +  3S  + 

3SO2 
Likewise  the  reaction: 

(339)  2AgNO3  +  2KOH  =  2KNO3  +  H2O  +  Ag2O 
is  composed  of  a  metathesis  and  decomposition: 

(340)  2AgN03  +  2KOH  =  2KNO3  +  2AgOH 

(341)  2AgOH  =  Ag2O  +  H2O 

Other  reactions  of  this  type  may  involve  amphoteric 
substances,  as,  for  example: 

(342)  A1C13  +  3NaOH  =  Al(OH),  +  3NaCl    (metathe- 

sis) 

(343)  A1(OH)3  =  H3A1O3  (amphoteric  substance) 

(344)  H3A1O3  +  3NaOH  =  Na3AlO3  +  3H2O 

(neutralization) 

Adding  these  three  equations  together  and  cancelling  the 
intermediate  products  gives: 

(345)  A1C13  +  6NaOH  =  3NaCl  +  Na3AlO3  +  3H2O 
Similar  equations  are: 

(346)  Pb(NO3)2  +  2KOH  =  Pb(OH)2  +  2KNO3 

(347)  Pb(OH)2  =  H2PbO2 

(348)  H2PbO2  +  2KOH  =  K2PbO2  +  2H2O 

(349)  Pb(NO3)2  +  4KOH  =  2KNO3  +  K2PbO2  + 

2H20 
and 

(360)  SnCl2  +  2NaOH  =  2NaCl  +  Sn(OH)2 
(351)  Sn(OH)2  =  H2SnO2 


98  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

(362)  H2Sn02  +  2NaOH  =  Na2SnO2  +  2H2O 

(353)  SnCl2  +  4NaOH  =  2NaCl  +  Na2Sn02  +  2H2O 

In  a  similar  way  some  sulfides  are  formed : 

(354)  3HgCl2  +  2H2S  =  4HC1  +  2HgS.HgCl2 

(355)  2HgS.HgCl2  +  H2S  =  2HC1  +  3HgS 

(356)  3HgCl2  +  3H2S  =  6HC1  +  3HgS 

and 

(357)  2PbCl2  +  H2S  =  2HC1  +  PbS.PbCl2 

(358)  PbS.PbCl2  +  H2S  -  2HC1  +  2PbS 

(359)  2PbCl2  +  2H2S  =  4HC1  +  2PbS 

It  may  be  that  only  a  part  of  a  double  salt  will  react, 
and  the  resulting  equation  will  appear  complex,  e.g. 

(360)  2(NH4)3AsS3  +  6HC1  =  6NH4C1  +  As2S3  +  3H2S 

However,  this  equation  becomes  simple  if  the  double  salt 
is  considered  as  composed  of  two  salts,  e.g. 

(361)  2(NH4)3AsS3  =  As2S3.3(NH4)2S 
and  one  of  these  reacts: 

(362)  3(NH4)2S  +  6HC1  =  6NH4C1  +  3H2S 
Other  examples  are 

(363)  2KMnO4  +  3H2SO4  +  5H2O2  =  K2SO4  + 

2MnS04  +  8H2O  +  5O2 

which  is  composed  of  the  three  steps: 

(364)  2KMn04  +  H2SO4  =  2HMnO4  +  K2SO4 

(365)  2HMnO4  +  5H2O2  =  2Mn(OH)2  +  4H2O  -  5O2 

(366)  2Mn(OH)2  +  2H2SO4  =  2MnSO4  +  4H2O 

or  again: 

(367)  2AuCl3  +  3H2O2  +  6NaOH  =  2Au  +  6NaCl  + 

3O2  +  6H20 


TYPES  OF  REACTIONS  AND  THEIR  EQUATIONS        99 

which  consists  of  the  three  reactions: 

(368)  2AuCl3  +  GNaOH  =  2Au(OH)3  +  GNaCl 

(369)  2Au(OH)3  =  Au2O3  +  3H2O 

(370)  Au2O3  +  3H2O2  =  2Au  +  3O2  +  3H20 

In  the  following  complex  reactions  two  or  more  steps 
can  be  recognized,  and  accordingly  two  or  more  equations 
should  be  written  for  every  case: 

(371)  MnO2  +  4HC1  =  MnCl2  +  2H2O  +  C12 

(372)  2BaCl2  +  K2Cr2O7  -  H20  =  2BaCr04  +  2KC1 

+  2HC1 

(373)  MnCl2  +  2KOH  -  H2O2  =  2KC1  +  H2O  + 

MnO(OH)2 

(374)  3NH3  +  3NaOCl  =  3NaCl  +  NH4OH  +  H2O 

When  problems  involve  a  series  of  incomplete  oxida- 
tions and  reductions,  it  is  well  to  construct  separate 
equations  for  each  reaction.  An  example  of  this  type  is 
offered  by  the  following  experiment:  By  adding  sulfuric 
acid  to  potassium  iodide  a  number  of  reaction  products 
are  formed.  These  reaction  products  are  hydroiodic 
acid,  iodine,  sulfur,  sulfur  dioxide,  and  hydrogen  sulfide 
besides  potassium  sulfate  and  water.  It  is  impossible  to 
see  from  a  single  equation  the  changes  which  have 
occurred,  and  each  oxidation  and  reduction  must  be 
treated  separately: 

A.  The  formation  of  hydroiodic  acid: 

(375)  2KI  +  H2SO4  =  K2SO4  +  HI 

B.  The  formation  of  sulfur  dioxide  and  iodine: 

(376)  2KI  +  2H2SO4  =  K2SO4  +  21  +  SO2  +  2H2O 

C.  The  formation  of  sulfur  and  iodine : 

(377)  6KI  +  4H2SO4  =  3K2SO4  +  61  +  S  +  4H2O 

D.  The  formation  of  hydrogen  sulfide  and  iodine: 

(378)  SKI  +  5H2S04  -  4K2S04  +  81  +  H2S  +  2H2O 


100  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

These  equations  show  that  in  A  there  is  a  simple  meta- 
thesis, the  replacement  of  a  weaker  acid  by  a  stronger 
acid;  in  B  the  iodide  is  oxidized  to  iodine,  and  the  hexa- 
valent  sulfur  is  reduced  to  tetravalent  sulfur,  in  C  to  free 
sulfur,  and  finally  in  D  to  sulfide.  It  is  quite  possible 
th&t  the  reactions  B,  C,  and  D,  are  secondary  reactions, 
as  the  HI  formed  in  reaction  A  may  be  decomposed  by 
the  sulfuric  acid.  If  such  is  the  case  the  KI  of  the  equa- 
tions should  be  replaced  by  HI,  and  the  unfinished  equa- 
tions are: 

(6)  HI  +  H+  +  S04—  =  I  +  S02  +  H20 

(c)  HI  +  H+  +  S04—  =  I  +  S  +  H20 

(d)  HI  +  H+  +  S04—  =  I  +  H2S  +  H20 

If  the  equations  (375),  (376),  (377)  and  (378)  are  added 
the  following  "monster"  equation  results: 

(379)   18KI  +  12H2S04  =  9K2SO4  +  2HI  +  161  + 

SO2  +  S  +  H2S  +  10H2O 

QUESTIONS  AND  PROBLEMS 

1.  Classify  the  following  reactions  according  to  type  and  note  partic- 
ularly whether  oxidation  and  reduction  is  involved.  If  so,  which  ele- 
ments are  oxidized  and  which  are  reduced? 

(380)  AgNO3  4-  NaNO2  =  AgNO2  +  NaNO3 

(381)  NaCl  +  AgOH  =  NaOH  +  AgCl 

(382)  2Hg+  +  Cr04~  =  Hg2CrO4 

(383)  Fe(OH),  +  3HBr  =  FeBr3  +  3H2O 

(384)  PtS  +  H,  =  Pt  +  H2S 

(385)  MnCl2  +  Na2CO3  =  MnCO3  +  2NaCl 

(386)  2K2MnO4  +  H2O  =  2KOH  +  K2Mn2O6  +  O2 

(387)  Mn++  +  2C1-  +  4H2O  =  MnCl2.4H2O 

(388)  CdCl2  H-  4RbCl  =  Rb4CdCl6 

(389)  Hg++  +  Hg  =  2Hg+ 

(390)  BeO  +  H20  =  Be(OH)2 

(391)  T1+  -f  Cl-  =  T1C1 

(392)  T1NO3  +  NaCl  =  T1C1  +  NaNO3 

(393)  2MnO2  +  2H2S04  =  2MnSO4  +  2H2O  +  O2 
(396)  TiCl4  +  4H20  =  H4TiO4  +  4HC1 

(396)  Th(N03)4  -  Th02  +  N2O6 


TYPES   OF   REACTIONS    AND   THEIR   EQUATIONS  101 

(397)  KMgCl3  =  KC1  +  MgCl2 

(398)  KOH  +  H2S  =  H2O  +  KSH 

(399)  SiO2  +  2CaF2  -f  H2SO4  =  2CaSO4  +  SiF4  +  2H2O 

(400)  CH4  +  2O2  =  CO2  +  2H2O 

(401)  CS2  +  3O2  =  CO2  +  2SO2 

(402)  B2S2  +  6H2O  =  2B(OH)3  +  3H2S 

(403)  H3BO3  =  HBO2  +  H2O 

(404)  2HBO2  =  B2O3  -f  H2O 

(405)  NaH2PO4  =  NaPO3  +  H2O 

(406)  2Na2HPO4  =  Na2P2O7  +  H2O 

(407)  BaO2  +  2HC1  =  BaCl2  +  H2O2 

(408)  PbO2  +  4HC1  =  PbCl4  +  2H2O 

(409)  MnO2  +  4HC1  =  MnCl4   +  2H2O 

(410)  PbCl4  =  C12  +  PbCl2 

(411)  MnCl4  =  da  +  MnCl2 

(412)  2HBr  +  H2SO4  =  Br2  +  SO2  +  2H2O 

(413)  KBr  +  KC103  =  KC1  +  KBrO3 

(414)  21  +  H2S  =  2HI  -h  S 

(415)  2HC1  +  F2  =  2HF  +  C12 

(416)  2AsH3  +  3Na  =  2Na3As  +  3H2 

(417)  2H2S  +  SO2  =  3S  +  2H2O 

(418)  6F  +  3H2O  =  6HF  +  O3 

(419)  2KI  +  O3  +  H2O  =  2KOH  +  I2  +  O2 

(420)  As4O6  +  3C  =  As4  +  3COi 

(421)  4SbCl3  -f-  6H2O  =  Sb4O6  +  12HC1 

(422)  2Ga(NO3)3  =  Ga2O3  +  3N2O5 

(423)  NaNOj  +  Pb  =  NaNO2  +  PbO 

(424)  A12(C03)3  =  A1203  +  3C02 

(425)  A12(S04)3  +  K2S04  =  2KA1(SO4)2 

(426)  A1(OH)8  =  HA1O2  +  H2O 

(427)  ZnSO4.7H2O  =  ZnSO4  +  7H2O 

(428)  ZnSO4.7H2O  =  Zn++  +  SO4~  +  7H2O 

(429)  ZnSO4  =  Zn~  +  SO4~ 

(430)  Co(OOC)2  =  Co  +  2C02 

(431)  CoCO3  =  CoO  +  CO2 

(432)  2Cr(OH)3  +  3H2SO4  =  Cr2(SO4)3  +  3H2O  . 

(433)  Cr04~  -  Pb++  =  PbCr04 

(434)  Pb++  +    21-  =  PbI2 

(435)  PbCl4  +  2H2O  =  Pb02  +  4HC1 

(436)  Te  +  SO8  =  TeO  +  SO2 

(437)  HAuCl4  =  AuCl  +  C12  -h  HC1 

(438)  HAuCU  =  And,  +  HC1 

(439)  Ag2O  +  H2O  =  2AgOH 

(440)  Ag+  -f  OH-  =  AgOH 


102  CHEMICAL  REACTIONS  AND  THEIR  EQUATIONS 

(441)  2AgOH  =  Ag20  +  H2O 

(442)  2A1  +  6HC1  =  2A1C1,  +  3H2 

(443)  2A1C1,  =  2A1  +  3C12 

(444)  AlCla  +  3H2O  =  A1(OH)3  +  3HC1 
(446)  Al(OH),  +  3HC1  =  A1C13  +  3H2O 
(446)  2A1  +  3S  =  A12S3 

2.  Write  reactions  (380)  and  (381)  in  the  ionic  form  and  deduce  their 
general  meaning  and  the  experimental  conditions. 

3.  Devise  a  laboratory  experiment  to  prove  the  correctness  of  reactions 
(382),  (389),  (433),  and  434. 

4.  Translate  the  meaning  of  reaction  (387)  into  common  language. 

6.  What  is  the  difference  between  the  reactions  (407),  (408)  and  (409)? 
What  equation  results  if  *(408)  is  added  to  (410),  and  (409)  is  added  to 
(411)? 

6.  Can    you  predict  from  the  displacement  series  that  the  reactions 
shown  in  equations  (414),  (415),  (418),  (420),  (384),  and  (416)  will  take 
place?     On  what  grounds  do  you  base  your  predictions? 

7.  What  is  the  meaning  of  equations  (427),  (428)  and  (429),  and  what 
actual  phenomena  or  reactions  correspond  to  these  equations? 

8.  Is  there  an  essential  difference  between  the  reactions  expressed  in 
(430)  and  (431)? 

9.  What  is  the  difference  or  similarity  between  (431)  and  (422),  (426) 
and  (427),  (396)  and  (403)  and  (404)? 

10.  Compare  equations  (387)  and  (428)  and  state  if  either  of  these 
reactions  is  a  reversible  one? 

11.  What  will  happen  if  Br2  is  substituted  for  F2  in  reaction   (415)? 
What  will  happen  if  HBr  is  taken  instead  of  HC1? 

12.  Write  the  complex  reaction  obtained  by  adding  equations  (437) 
and  (438)  together.     Describe  the  experimental  conditions. 

13.  What  is  the  meaning  of  equations  (439),  (440),  and  (441)?     Devise 
experiments  for  each  reaction. 

14.  State  in  common  terms  the  information  contained  in  equations 
(442),    (443),  (444),  (445).     Suggest  for  each  equation  the  experimental 
conditions  under  which  the  reaction  takes  place. 

16.  How  will  the  speed  of  reaction  (446)  be  affected  if  S  is  replaced  by 
(a)  Se,  (6)  Te,  (c)  O? 

16.  Why  does  the  Te  in  (436)  rob  the  S  of  a  part  of  its  O?     Suggest  an 
explanation. 

17.  What  is  the  difference  between  equations  (391)  and  (392)? 

18.  What  takes  place  in  the  reactions  expressed  in  equations  (400)  and 
(401)?     Give  a  common  term  for  the  phenomena. 

19.  Does  S  replace  O,  or  does  O  replace  S  in  reaction  (402)  ?     Is  this 
reaction  in  harmony  with  the  displacement  series? 


APPENDIX  I 

KEY  TO  NOMENCLATURE  OF  CHEMICAL 
COMPOUNDS 

This  key  to  the  nomenclature  of  inorganic  compounds 
enables  the  beginner  to  construct  numerous  formulas 
of  compounds  whose  names  are  given,  and  likewise  to 
find  the  correct  name  for  a  given  formula.  It  will  also 
aid  in  finding  possible  reaction  products  in  cases  where  a 
precipitate  or  a  change  of  color  occurs  during  the  reaction. 

To  construct  the  formula  for  a  compound  it  is  only 
necessary  to  join  the  radicals  or  ions  in  such  a  way 
that  all  the  valencies  or  charges  are  satisfied  or  neutral- 
ized. The  formula  for  "cupric  sulfate"  is  found  by 
joining  Cu++  (cupric  ion)  and  SO4 —  (sulfate  ion)  to 
CuSO4.  The  formula  for  "  ferric  chromate"  is  found  by 
joining  2Fe+++  (=  six  positive  charges)  with  3Cr04 — 
(six  negative  charges)  to  Fe2(Cr04)3.  Likewise  cupric 
chlorate  requires  Cu++  and  2C1O3~,  that  is  Cu(ClO3)2. 
Cuprous  phosphide  is  Cu3P,  and  cupric  phosphide  is 
Cu3P2. 

By  finding  the  ion  or  radical  under  the  formulas  the 
name  of  a  given  formula  can  be  constructed :  FeS  shows 
iron  to  be  bivalent  (Fe^*),  hence  ferrous  sulfide,  while 
in  Fe2S3  the  iron  is  trivalent  (Fe+++),  hence  ferric 
sulfide.  Likewise  the  formulas  Nad,  NaClO2,  NaClO3, 
NaC104  represent  respectively  sodium  chloride,  sodium 
chlorate,  sodium  chlorate,  and  sodium  perchlorate. 

The  column  indicating  the  color  of  the  ion  or  radical 
is  of  advantage  in  laboratory  work  where  the  formation 

103 


104 


APPENDIX 


Name  of  element,  ions  or 
radicals 


Valence 
number 


Symbol  of 


Element 


Cation 


Anion 


Color  of 
ions 


Aluminum 0 

aluminum  ion 

aluminate  ion 

Ammonia 0 

ammonium 1 

Antimony 0 

antimonous 

antimonic 5 

antimoniate 5 

antimonite 3 

Arsenic 0 

arsenpus 3 

arsenic 5 

arsenide 

arsenite 

arsenate 5 

Barium 0 

barium 2 

Bismuth 0 

bismuth 

bismuthyl 

Boron 0 

boron 3 

borate 

tetraborate 3 

Bromine 0 

bromide —  1 

bromite 

bromate 5 

Cadmium 0 

cadmium 2 

Calcium 0 

calcium 2 

Carbon 0 

carbide 4 

carbonate 4 

bicarbonate 4 

Cerium 0 

cerous 3 

eerie 4 

Chlorine 0 

chloride —1 

hypochlorite 1 

chlorite 3 

chlorate 5 

perchlorate 7 

Chromium 0 

chrompus 2 

chromic 3 

chromite 3 

chromate 5 

bichromate 5 

Cobalt 0 

cobaltous 2 

cobaltic 3 

Copper 0 

cuprous 1 

cupric 2 

copper  ammonium 2 


Al 

NHa 
Sb 


As 

Ba 
Bi 
B 

Br 

Cd 
Ca 
C 

Ce 
Cl 

Cr 

Co 
Cu 


AS+ 


BiO+ 

B+++ 


Cd++ 
Ca++ 


Ce+ 


A102- 


SbO?- 


Cs'" 
AsOs 
As04 


B03" 


Br~ 
|Br02- 
BrOr 


C" 

co3~- 

HC03- 


ci- 

cio- 

cio2- 

ClOj- 


Cr02- 
CrOs- 
Cr20; 


Co++ 
Co+++ 

Cu+ 

Cu++ 

Cu(NH8)4 


blue 

purple 

green 

yellow 

orange 

red 
(unstable) 

colorless 
blue 
deep  blue 


APPENDIX 


105 


Name  of  element,  ions,  or 
radicals 


Valence 
number 


Symbol  of 


Element        Cation 


Anion 


Color  of 
ions 


Fluorine 0 

fluoride - 1 

Gold 0 

aurpus 

auric 

Hydrogen 

hydrogen 1 

hydride -1 

Iodine 0 

iodide ~  1 

iodate 5 

periodate 7 

Iron 

ferrous 

ferric 

ferrate 6 

Lead 0 

plumbpus 

plumbic 4 

plumbate 4 

Lithium 0 

lithium 

Magnesium 

magnesium    2 

Manganese 

manganous 

manganic 3 

manganate 5 

permanganate 7 

Mercury 0 

mercurpus 1 

mercuric.  ...  2 

Nickel 0 

niccolpus 2 

niccolic 

Nitrogen 0 

nitride —  3 

nitrite 3 

nitrate 5 

Oxygen 0 

oxide -2 

Phosphorous 0 

Ehosphide —  3 

ypophosphite 1 

phosphite 3 

phosphate 5 

metaphosphate 5 

monophosphate 5 

pyrophosphate 5 

Platinum 0 

platinpus 2 

platinic 4 

Potassium 0 

potassium 1 


F 
Au 

H 
I 

Fe 

Pb 

Li 

Mg 
Mn 

Hg 

Ni 
N 

.O 

P 


Pt 
K 


Au++ 
H+ 


Fe++ 
Fe+++ 


Pb+ 
Pb+ 


Mg+ 
Mn+ 


Hg+ 


Pt+ 


F- 


H2 


I- 

ror 


Fe04~ 
Pb08- 


MnO*~ 
MnO4 


N"1 

NO2- 

N03- 

0" 
pur 

H2P02- 
H8P03 
POi 

PO,- 

H2P04 


yellow 
yellow 


green 
yellow- 
brown 
red 


si.  pink 
si.  green 
green 
purple 


green 
(unstable) 


yellow 
orange 


106 


APPENDIX 


Name  of  element,  ions  or 

Valence 

Symbol  of 

Color  of 

radicals 

number 

Element 

Cation 

Anion 

ions 

Silicon     .    . 

o 

Si 

silicic!  e  .• 

—  4 

SiIV 

silicate           

4 

SiO4  

metasilicate  

4 

SiOs  — 

Silver                          .      ... 

0 

Aff 

silver  

1 

Ag+ 

Sodium 

0 

Na 

sodium.          

1 

Na+ 

Strontium 

0 

Sr 

strontium  

2 

Sr++ 

Sulphur 

o 

g 

sulphide  
hydrosulnde  

-2 
-2 

s— 

HS- 

sulfite 

4 

SOs  — 

bisulfite  

4 

HSO3~ 

sulfate 

6 

SO4  — 

bisulf  ate               

6 

HSO4~ 

Tin 
stannous  

0 
2 

Sn 

Sn++ 

stannite  

2 

SnOs  — 

stannic  .  .         

4 

Sn++++ 

stannate  

4 

SnOa  — 

Zinc     

o 

Zn 

zinc  

2 

Zn++ 

zincate          .          

2 

ZnO2 

zinc  ammonium  

2 

Zn(NHa)4++ 

Uranium 

0 

u 

uranous      

2 

U++ 

light  green 

uranyl 

6 

UOa++ 

yellow 

uranate  

6 

UOi" 

yellow 

Common  Organic  Radicals 
acetates  (CHsCOQ-)  ..... 
cyanides  

1 
1 

Ac- 

CN- 

cyanates  

1 

CNO- 

ferrocyanides  

4 

Fe(CN)6 

ferricyanides              .... 

3 

Fe(CN)e 

2 

Ox~ 

rhodanates                  .... 

1 

CNS- 

of  a  distinctive  color  in  the  test  tube  will  show  the 
respective  ion  or  compound.  Thus,  if  during  a  chemical 
reaction,  the  yellow  solution  of  a  chromate  turns  green- 
chromic  ion  has  been  formed,  if  the  purple  permanganate 
solution  turns  green — manganate  ion  has  been  produced, 
and  so  on.  Where  no  color  is  given  the  substance  is 
colorless  or  white. 

If  necessary  the  student  may  enlarge  the  list  to  meet 
special  requirements.  A  frequent  use  of  this  key  in  the 
laboratory  is  recommended. 


APPENDIX  II 
DISPLACEMENT  SERIES 

(Elements  printed  in  capitals  to  be  memorized  by  the 

student) 

In  the  displacement  series  the  elements  are  arranged 
according  to  their  electro-motive  force,  that  is,  the 
capacity  of  holding  the  ionic  charges  more  or  less  firmly. 

NEGATIVE  tantalum                        columbium 

palladium  cadmium 

fluorine  ruthenium  IRON 

CHLORINE  rhodium  ZINC 

OXYGEN  ANTIMONY                    manganese 

NITROGEN  BISMUTH                       uranium 

BROMINE  ARSENIC                         gadolinium 

IODINE  MERCURY                      indium 

SULFUR  SILVER                            gallium 

selenium  COPPER  ALLUMINIUM 

tellurium  silicon                             rare-earth  metals 

PHOSPHORUS  titanium                         beryllium 
chromium                                                                    scandium 

vanadium  HYDROGEN                   yttrium 

tungsten  MAGNESIUM 

molybdenum  TIN                                     lithium 

CARBON  LEAD  CALCIUM 

boron  germanium                     strontium 

GOLD  zirconium  BARIUM 

osmium  cerium  SODIUM 

platinum  nickel  POTASSIUM 

iridium  cobalt 

thallium  POSITIVE 
107 


108  APPENDIX 

At  the  top  of  the  list  are  the  most  negative  elements 
which  displace  all  following  anions.  The  elements  near 
the  end  of  the  list  are  the  most  positive  and  will  displace 
each  cation  above  them.  Thus  Br  will  displace  I  and  S, 
but  F  will  displace  Cl  and  Br.  Ca  will  displace  Mg,  Al, 
and  Fe,  but  K  will  displace  Na  and  Ca,  etc. 

The  farther  apart  the  elements,  the  more  stable  their 
compounds.  Thus  KF  is  the  most  stable  compound, 
while  a  compound  of  0  and  F  is  so  unstable  that  it 
cannot  exist. 


APPENDIX  III 

THE    PERIODIC    SYSTEM    AND    THE 
CLASSIFICATION  OF  THE  ELEMENTS 

A  systematical  arrangement  of  all  chemical  elements 
is  essential  for  an  understanding  of  their  correlation, 
their  similarity  or  difference.  The  periodic  system  is 
such  a  classification  in  which  all  elements  are  logically 
tabulated.  The  position  occupied  by  an  element  in  this 
system  is  an  indication  of  its  chemical  and  physical 
properties. 

The  periodic  system  was  not  suddenly  discovered  as  is 
often  stated,  but  has  gradually  developed  and  taken 
form  with  increasing  chemical  knowledge  during  the 
last  100  years.  One  of  the  earliest  attempts  at  classifi- 
cation was  made  in  1829  by  Doebereiner  who  arranged 
some  elements  in  triads  or  groups-of-three,  such  as 
Ca-Sr-Ba,  Li-Na-K,  Cl-Br-I,  etc.;  because  the  properties 
and  the  atomic  weight  in  each  triad  show  a  certain 
relationship.  The  same  idea  was  further  developed  by 
Cooke  in  1854  and  by  DeChancourtois  in  1865.  The 
next  noteworthy  step  was  taken  in  1867  by  Newlands  in 
his  octaves  or  groups-of-eight  in  which  he  pointed  out, 
that  if  all  elements  are  arranged  according  to  increasing 
atomic  weights,  each  eighth  element  has  analogous 
properties,  for  elements  of  similar  properties  are  sepa- 
rated by  seven  others,  just  as  is  the  case  in  a  harmonious 
musical  scale.  This  law  of  octaves  contained  the  nucleus 
of  the  periodic  system  and  in  1869  Mendeleeff  and 
Lothar  Meyer,  working  independently  of  each  other, 

109 


110  APPENDIX 

presented  the  first  table  of  a  system  of  elements  which 
was  generally  accepted.  •  Mendeleeff  predicted  the 
properties  of  three  unknown  elements  from  three  empty 
spaces  occuring  in  his  table,  and  when  these  elements 
were  later  discovered,  it  was  found  that  his  predictions 
came  true  with  an  astounding  accuracy.  Since  that 
time  numerous  discoveries  have  been  made  and  the 
system  has  gradually  been  completed.  Today  we  may 
assume  with  great  confidence  that  only  five  elements 
remain  undiscovered  and  that  the  total  number  of 
elements  in  the  series  from  hydrogen  to  uranium  inclusive 
is  92.  The  five  unknown  elements  with  atomic  numbers 
43,  61,  75,  85,  and  87  would  complete  the  table. 

If  all  elements  are  arranged  in  the  order  of  increasing 
atomic  weights,  then  the  92  elements  between  H  (At.  No. 
1)  and  U  (At.  No.  92)  inclusive  are  divided  into  six  periods, 
the  terminals  of  each  period  being  the  noble  gases.  Thus 
there  are  in : 

Period      I,  extending  from  He  to  Ne  8  elements  ( =  2  X  22) 

Period    II,  extending  from  Ne  to  Ar  8  elements  ( =  2  X  22) 

Period  III,  extending  from  Ar  to  Kr  18  elements  ( =  2  X  32) 

Period  IV,  extending  from  Kr  to  Xe  18  elements  ( =  2  X  32) 

Period    V,  extending  from  Xe  to  Nt  32  elements  (=  2  X  42) 

Period  VI,  extending  from  Nt  to  U  7  elements 

Each  period  has  the  following  characteristics : 

The  terminals  of  each  period  are  electrically  neutral 
(noble  gases). 

The  first  three  elements  of  each  period  are  strongly 
electropositive  (light  metals). 

The  last  three  elements  of  each  period  are  strongly 
electronegative  (non-metals). 

The  central  membets  of  each  period  are  amphoteric 
(heavy  metals). 

The  members  of  the  carbon  family  are  the  framework 


APPENDIX 


111 


of  the  system  and  form  a  transition  between  the  different 
groups  of  elements  (C  the  life-element,  Si  the  rock- 
forming  element). 

Besides  the  horizontal  division  into  periods  there  is 
the  vertical  division  into  groups.  The  general  rule  of 
the  periodic  chart  is : 

The  similarity  among  the  elements  in  the  upper  half 


The  Periodic  System 


NON-METALS  INERT  ELEMENTS 

Group  4        5a      6a      7a  0  la 


111! It  If 


LIGHT  MEf  ALS 

2a      3a      4a 


-ft  £  a 


Period 
Vb 


IVb 

Illb 

lib 

Ib 


Iron  period 
III' 


Silver  period 


er  pi 
IV 


Rare  earth 
metals 

V" 

Gold 
period 

Radioactive 

elements 

VI  ' 

Group 


82 
Pb 

83 
Bi 

84 
Po 

85 

86 

Nt 

87 

88 
Ra 

89 
Ac 

90 
Th 

50 

Sn 

51 
Sb 

52 
Te 

53 

54 

Xe 

55 
Cs 

56 
Ba 

57 
La 

58 
Ce 

32 
Ge 

33 

As 

34 

Se 

35 

Br 

36 
Kr 

37 
Rb 

38 

Sr 

39 
Y 

40 
Zr 

14 
Si 

15 
P 

16 
S 

17 
Cl 

18 
Ar 

19 
K 

20 
Ca 

21 

Sc 

22 
Ti 

6 
C 

N 

8 
0 

9 
F 

10 

Ne 

2 
He 

11 
Na 

3 
Li 

12 
Mg 

4 
Be 

13 
Al 

5 
B 

14 
Si 

6 
C 

H 

22      23      24      25 
Ti      V      Cr     Mn 

26     27    28 
Fe    Co    Ni 

29      30      31      32 
Cu     Zn    Ga    Ge 

40     41       42      43 
Zr    Cb     Mo 

44     45     46 
Ru   Rh   Pd 

47     48      49      50 
Ag     Cd     In       Sn 

58  59  60  61  62  63  64  65  66  67  68  69    70    71  72 
Ce  Pr  Nd      Sm  Eu  Gd  Tb  Dy  Ho  Er  Db  Tm  Yb  Lu 

72      73  74      75 
Lu     Ta    W 

76     77     78 
Os     Ir      Pt 

79      80      81     82 
Au     Hg     Tl    Pb 

90      91      92 
Th     Bv       U 

VI 

Va 

IVa 

Ilia 

Ila 

la 

Iron  period 
III' 


Silver 

period 

IV 

Rare  earth 

metals 

V" 

Gold 
period 


5b      6b      7b         888 
HEAVY  METALS 


Ib      2b      3b 


112 


APPENDIX 


of  the  table  is  most  pronounced  in  the  vertical  direction, 
therefore  these  elements  possess  group-analogy. 

The  similarity  among  the  elements  of  the  lower  half 
of  the  table  is  most  pronounced  in  the  horizontal  direct- 
ion, therefore  these  elements  possess  period-analogy. 

ATOMIC  NUMBERS  AND  ATOMIC  WEIGHTS 

Importance  of  the  elements  is  indicated  by  *  and  o.     The  atomic  numbers  serve  as 
n  index  to  the  periodic  table. 


Actinium 

89 
13 
51 
18 
33 

56 
4 
83 
5 

,91 

35 

48 
55 
20 
6 

58 
17 
24 
27 
41 

29 
69 
66 
68 
63 

9 
64 
31 
32 
79 

2 
1 
49 
53 

77 

26 
36 
57 

82 
3 

72 
12 
25 
80 
42 

Ac 
Al 
Sb 
A 
As 

Ba 
Be 
Bi 
B 
Bv 

Br 
Cd 
Cs 
Ca 
C 

Ce 
Cl 
Cr 
Co 
Cb 

Cu 
Db 
Dy 
Er 
Eu 

F 
Gd 
Ga 
Ge 
Au 

He 
H 
In 
I 
Ir 

Fe 
Kr 
La 
Pb 
Li 

Lu 

Mg 
Mn 
Hg 
Mo 

27.1 
120.2 
39.88 
74.96 

137.37 
9.1 
208.0 
11.0 

79.92 
112.4 
132.81 
40.07 
12.00 

140  .25 
35.46 
52.0 
58.97 
93.1 

63.57 
168.5 
162.5 
167.5 
152.0 

19.0 
157.3 
69.9. 
72.5 
197.2 

4.00 
1.008 
114.8 
126.96 
193.1 

55.84 
82.92 
139.0 
207.2 
6.49 

17.50 
24.32 
54.93 
200.6 
96.0 

Neodymium  .... 

60 
10 
28 
7 
76 

8 
46 
15 

78 
84 

19 
59 
88 
45 
37 

44 
62 
21 
34 
14 

47 
11 
38 
16 
73 

52 
65 
81 
90 
70 

50 
22 
74 
92 
23 

54 
71 
39 
30 
40 

tBeryl 
(Colun 
,isY. 
jod)  is  J 

Nd 

Ne 
Ni 
N 
Os 

0 
Pd 
P 
Pt 
Po 

K 
Pr 
Ra 
Rh 
Rb 

Ru 

Sm 
Sc 
Se 
Si 

Ag 
Na 
Sr 
S 
Ta 

Te 
Tb 
Tl 
Th 
Tm 

Sn 
Ti 
W 
U 
V 

X 

Yb 
Y 
Zn 
Zr 

ium) 
ibiuir 

. 

144.3 
20.2 
58.68 
14.01 
190.9 

16.000 
106.7 
31.04 
195.2 

39.10 
140.9 
226.0 
102.9 

85.45 

101.7 
150.4 
44.1 
79.2 
28.3 

107.88 
23.00 
87.63 
32.06 
181.5 

127.5 
159.2 
204.0 
232.4 
170.5 

118.7 
48.1 
184.0 
238.2 
51.0 

130.2 
173.5 
88.7 
65.37 
90.6 

r 

*  Aluminum  
Antimony  
Argon  
Arsenic  

Neon  
o  Nickel 

*  Nitrogen  
Osmium  

*  Oxygen  
Palladium  

o  Barium             .  . 

Beryllium  
o  Bismuth 

*  Phosphorus  
o  Platinum  

o  Boron          

Brevium 

Polonium  

*  Potassium  
Praseodymium..  . 
o  Radium       

o  Bromine           •    •  • 

Cadmium  
Caesium                .  . 

*  Calcium  
*  Carbon  

Rhodium   

Rubidium  

Ruthenium  
Samarium  
Scandium 

Cerium  
*  Chlorine     

o  Chromium 

o  Cobalt          

o  Selenium  
*  Silicon 

Columbium 

o  Copper              .... 

o  Silver           .      .  . 

Denebium       

*  Sodium       

Dysprosium     .... 

o  Strontium     

*  Sulfur  
Tantalum  

Tellurium 

Europium             .  . 

o  Fluorine 

Gadolinium  
Gallium   

Terbium  
Thallium  
o  Thorium  
Thulium  

Germanium  

Gold       

o  Helium  

o  Tin  

*  Hydrogen  
Indium  
.    Iodine  

o  Titanium  .....    . 
Tungsten  
o  Uranium  
Vanadium  

Xenon  

Iridium  
*Iron 

Krypton     .  .  . 

Ytterbium  
Yttrium 

Lanthanum  
o  Lead 

o  Zinc  

Lithium  

Lutecium  
*  Magnesium  
o  Manganese 

Zirconium  

Gl  (glucinum)  is  Be 
Nb  (niobium)  is  Cb 
Flis,F;  Xeis  X;  Y1 
Az  (azote)  is  N  ;  J  ( 

o  Mercury 

Molybdenum  

APPENDIX  113 

Accordingly  there  are  group-relations  and  period-rela- 
tions. E.g.  the  group-relation  of  Au  refers  to  its  proper- 
ties which  are  similar  to  those  of  Ag  and  Cu,  while  the 
period-relation  of  Au  refers  to  those  properties  which 
resemble  Hg  and  Pt;  in  other  words,  gold  has  a  group 
similarity  to  silver  and  copper,  and  a  period  similarity 
to  mercury  and  platinum. 

OUTLINE   OF  PROPERTIES  OF  ELEMENTS  AS  DEDUCED 
FROM  THE  PERIODIC  TABLE 

In  the  UPPER  half  of  the  table  are  the  elements  of 
the  strongest  e.m.f.,  the  simplest  spectra,  low  density, 
and  colorless  ions. 

In  the  LOWER  half  of  the  chart  are  the  elements  of 
the  weakest  e.m.f.,  complex  spectra,  high  density, 
and  colored  ions. 

On  the  LEFT  of  the  table  are  the  electro-negative 
elements  which  form  the  acids.  The  solid  elements  are 
brittle  and  paramagnetic. 

On  the  RIGHT  of  the  table  are  the  electro-positive 
elements  forming  the  bases.  The  solid  elements  are 
malleable. 

In  the  LEFT  UPPER  HALF  and  RIGHT  LOWER 
HALF  are  located  the  non-metals  and  heavy-metals 
which  are  found  in  their  native  form  in  nature  (free 
state).  Their  sulfides  are  common  and  they  are  readily 
reduced  to  the  free  state. 

In  the  RIGHT  UPPER  HALF  and  LEFT  LOWER 
HALF  are  the  light  metals  and  heavy  metals  which 
always  occur  in  combination  and  never  in  their  free 
state  in  nature.  Their  oxides  and  oxysalts  are  common, 
and  it  is  with  difficulty  that  they  are  reduced  to^free 
elements.  Not  all  elements  are  of  equal  importance. 
Fourteen  elements  are  of  primary  importance  (marked 


114  APPENDIX 

with  bold  type  and  *)  for  they  constitute  99  per  cent  of 
the  known  earth  surface  and  are  abundant  in  the  litho- 
sphere,  hydrosphere,  atmosphere.  These  elements  are 
essential  to  all  living  matter  (biosphere)  and  are  of 
fundamental  importance  in  every  industry.  About  40 
elements  are  of  secondary  importance  (marked  o),  for 
they  are  less  abundant  and  are  used  only  for  specific 
purposes  in  industry  and  technic.  About  33  elements  are 
of  more  or  less  scientific  interest  only  for  they  very  rarely 
occur  and  are  of  little  or  no  present  use. 

The  student  must  be  familiar  with  the  fourteen  im- 
portant and  fundamental  elements,  know  their  position 
in  the  table,  and  be  familiar  with  their  chemistry,  for 
they  are  most  common  on  the  surface  of  our  earth  and 
their  compounds  are  used  in  every  industry,  in  every 
material  science,  and  in  every-day  life.  Elementary 
chemistry  confines  itself  mainly  to  these  elements  as 
they  are  the  basis  for  a  study  of  the  other  elements. 
Thus,  e.g.  with  chlorine  as  type  or  example,  the  proper- 
ties of  fluorine,  bromine,  and  iodine  can  readily  be 
memorized;  while  with  sulfur  as  type  or  example  the 
characteristics  of  selenium  and  tellurium  are  easily 
learned;  potassium  and  sodium  is  the  type  for  lithium, 
rubidium,  and  cesium;  manganese  is  similar  to  chromium, 
iron  similar  to  cobalt  and  nickel — and  so  on. 

REFERENCES 

Classification  of  chemical  elements,  Scientific  American  Supplement, 
vol.  87,  p.  146,  1919. 

Modification  of  the  periodic  table,  Am.  Jour,  of  Science,  vol.  46,  p.  481, 
1918. 

Distribution  of  chemical  elements,  Science  Progress,  vol.  14,  p.  602, 
1920. 
BOOKS  ON  THE  SUBJECT: 

Rudorf,  G.,  The  Periodic  System. 

Garrett,  A.  E.,  The  Periodic  Law. 


APPENDIX  IV 

SOLUBILITY  TABLE  OF  COMPOUNDS 

(To  be  Memorized  by  Students) 

SOLUBLE  in  water: 

ALL  CHLORIDES,  except  AgCl,  PbCl2,  HgCl; 
ALL  NITRATES,  except  bismuth,  subnitrate; 
ALL  SULFATES,  except  barium  sulfate,  strontium 

sulfate,  calcium  sulfate,  lead  sulfate; 
ALL  ACETATES,  except  basic  ferric  acetate  and  basic 

aluminum  acetate ; 
ALL  ALKALI  SALTS,  except  acid  potassium  tartrate, 

acid  ammonium  tartrate,  potassium  platin  chloride, 

ammonium    platin    chloride,     and    sodium    pyro- 

antimoniate. 

INSOLUBLE  in  water: 

ALL  OXIDES  ] 

ATT  QTTT  T^TTvira  except  those  of  the  alkalies 

AL/lj  oU-LrlJJJiib  . 

ALL  HYDROXIDES  |  and  earth  alkali  metals  ; 

ALL  CARBONATES  ^ 

ATT  PHOSPHATFS       exceP^    those    01    the    alkali 

ALL  SILICATES  metals ; 

ALL  CHROMATES,    except    those  of    the   alkalies; 

Ca,  Sr. 

SOLUBILITY  TABLE  OF  COMPOUNDS 
,  (for  reference) 

ACETATES    are   SOLUBLE,    except   AgAc,    HgAc, 
HgAc2. 

115 


116 


APPENDIX 


ARSENITES  are  INSOLUBLE,  except  those  of  the 

alkalies. 
ARSENATES  are  INSOLUBLE,  except  those  of  the 

alkalies  and  the  acid  arsenates  of  the  earth  alkali 

metals. 

BORAXES  are  SOLUBLE. 
BROMATES     are     SOLUBLE,     least     soluble     are 

HgBrO,,  AgBr03. 
BROMIDES   are   SOLUBLE,    except   CuBr,    AgBr, 

HgBr,  PbBr2. 
CARBONATES  are  insoluble,  except  those  of  alkalies 

and  Tl. 

CHLORATES  are  SOLUBLE,  except  HgClO3. 
CHLORIDES  are  SOLUBLE,   except  AgCl,   HgCl; 

CuCl,  AuCl,  PtCl2,  PbCl2. 
CHROMATES   are   INSOLUBLE,   except   those   of 

alkalies  and  CaCr04,  SrCrO4  and  MgCrO4. 
CYANIDES  are  IN  SOLUBLE,  except  those  of  alkalies 

and  earth  alkalies. 
FERROCYANIDES  are  INSOLUBLE,  except  those 

of  alkalies  and  earth  alkalies. 
FERRICYANIDES  are  INSOLUBLE,  except  those  of 

alkalies  and  earth  alkalies. 
FLUORIDES    are    INSOLUBLE,    except    those    of 

alkalies  and  AgF,  FeF2,  SnF2,  SrF2  and  CdF2. 
IODATES  are  INSOLUBLE,  except  those  of  alkalies. 
IODIDES  are  SOLUBLE,   except  Agl,   Cul,   HgI2, 

PbI2,  Bil3,  SbI3,  Pt  metals. 
MANGANATES  are  INSOLUBLE,  except  those  of 

the  alkalies. 
MOLYBDATES  are  INSOLUBLE,  except  those  of 

the  alkalies. 

NITRATES  are  SOLUBLE. 
NITRITES  are  SOLUBLE,  except  AgNO2. 


APPENDIX  117 

OXALATES  are  INSOLUBLE,  except  those  of  the 

alkalies. 

PERMANGANATES  are  SOLUBLE,  except  AgMnO4. 
PHOSPHATES  are  INSOLUBLE,  except  those  of  the 

alkalies. 
PHOSPHITES  are  INSOLUBLE,  except  those  of  the 

alkalies,  and  Tl. 
SILICATES  are  INSOLUBLE,  except  those  of  the 

alkalies. 
SULFATES   are   SOLUBLE,    except   BaSO4,    SrSO4, 

PbS04,  slightly  soluble  are  Ca,  Ag,  Hg. 
SULFIDES  are  INSOLUBLE,   except  those  of  the 

alkalies. 


APPENDIX  V 
PREPARATION  OF  SALTS 

A  Metallic  Salt  May  be  Prepared  in  Any  One  of  the 
Following  Ways  (To  be  Memorized  by  the  Student) 

Reaction  between: 

1.  METAL  and  HALOGEN: 

Zn        +  C12          =  ZnCl2 

la.  METAL  and  ACID: 

Zn        H-   H2SO4  =  ZnSO4  +  H2 

2.  METAL-OXIDE  and  ACID-ANHYDRIDE : 

ZnO  +  SO3  =  ZnS04 

2a.  METAL-OXIDE  and  ACID: 

ZnO  +  H2S04  =  ZnSO4  +  H2O 

3.  METAL  HYDROXIDE  and  ACID  ANHYDRIDE: 

Zn(OH)2  +  SO3  =  ZnSO4  +  H2O 

3a.  METAL  HYDROXIDE  and  ACID: 

Zn(OH)2  +  H2SO4  =  ZnSO4  +  2H2O 

4.  METAL  CARBONATE  and  ACID: 

ZnCO3  +  H2S04  =  ZnSO4  +  CO2  +  H20 

5.  METAL  SULFIDE  and  ACID: 

ZnS  +  H2SO4  =  ZnS04  +  H2S 

6.  Interaction-of  two  METAL  SALTS: 

Zn(NO3)2  +  Na2SO4  =  ZnSO4  +  2NaNO3 

118 


APPENDIX  119 

KEY  TO  THE  EQUATIONS 

The  following  is  a  systematic  index  of  the  equations 
used  in  this  book.  The  numbers  to  the  left  of  the 
formula  refer  to  the  reacting  substances  (which  stand 
at  the  left  of  an  equation),  and  the  numbers  to  the 
right  refer  to  reaction  products  (which  stand  at  the 
right  side  of  an  equation). 

This  index  makes  it  possible  to  find  at  a  glance  how 
some  of  the  substances  are  prepared — thus  under  oxygen 
there  is  a  score  of  equations  showing  how  this  gas  may 
be  made.  It  also  gives  the  analytical  tests  for  many 
compounds,  that  is,  the  reactions  used  for  their  detection. 
Furthermore  it  will  facilitate  the  construction  of  similar 
equations  for  reactions  taking  place  with  related  ele- 
ments. Thus  many  reactions  given  under  sulfur  occur 
also  with  selenium  and  tellurium,  likewise  those  of 
arsenic  compounds  have  a  relation  to  those  of  antimony 
and  bismuth  compounds. 

It  is  well  for  students  to  make  a  comparative  study 
of  reactions  by  consulting  the  periodic  system  and  the 
displacement  series  and  extend  this  index  to  the  labora- 
tory and  lecture  note  book  for  further  reference. 

ALUMINUM 

442,  446 Al 443 

21,  338 Al+++         

162,  335,  337,  342,  345,  443,  A1C13         113-5,  442,  445 

444 

113 A12O3         424 

114 A12S3          127.446 

115 '  Al2Se3         

116 Al2Te3        

131 A14C3         

9,15,127,343,344,426,445.  A1(OH)3  20,  21,   131,   162,  335,  337 

338,  342,  444 

H3A103         343 

HA1O2 426 


120  APPENDIX 

,  A1(N03)3       9 

20,  425 A12(S04)3      

424 A12(C03)3 

A1P04         15 

AMMONIUM 

86,88,89,92,233,333,374..  NH3  101,     106,     168,     282,     294 

46 NH4+         

90,  91 NH4OH       46,  86,  374 

92,  101 NH4C1        88,  233,  333,  360,  362 

289 NH4N03      

94,  95,  96,  362 (NH4)2S       361 

360,  361 (NH4)2AsS3    94 

(NH4)2SnS3    95,  96 

289 (NH4)2MoO3 

ANTIMONY 

185 Sb            

165,  421 SbCl3         

Sb4O6         421 

80,  81 Sb2S3 

Sb2S5 150 

SbOCl         165 

ARSENIC 

186 As 420 

416 AsH3          118,  169 

420 As4O6         

94 . As2S3         321,  322,  360 

321. H3AsO3       272,  320 

320,  322 H3AsO4        

BARIUM 

18,  139,  140 Ba++          .  .  151 

17,  372 BaCl2         407 

BaO          243,  244,  246 

151,  407 BaO2         

157,  243 BaS           

BaC2          247 

7,  13 Ba(OH)2      157,  158 

158 Ba(SH)2       157 

Ba(N03)2      7 

243,^246 BaSO4        17,  18,  139 

244/247 BaCO8         


APPENDIX  121 

BaCrO4       ...    140,  372 

Ba3(PO4)3     .13 


BERYLLIUM 

390 BeO  

Be(OH)2  390 

BISMUTH 

184 Bi  

160 .' BiCl3  

'.'.          BiI3  184 

Bi(OH)3  161 

161 BiOCl  160 

BORON 

, B2O3 133,  404 

402 B2S2  

168 BN  106 

404 BO(OH)  168,  403 

133,  403 B(OH)3  402 

106 B(NH2)3  

BROMINE 

54,  213,  221,  263,  264,  285. ..  Br  27,  31,  32,  33,  53,   214,  220, 

412 

27,  33,  53 Br-  54 

31,  32,  383,  412 HBr  221,  263,  264 

CADMIUM 
388 CdCl2 


CALCIUM 

73,   177 Ca  70 

399 CaF2          123,  124 

75,  77 CaCl2  ....  .6,  76,  110,  111,  112,  177 

70,  74,  79,  84,  110,  120,  123.  CaO          69,  73 

CaS          120,  121 

112 Ca3P2         

169 Ca3As2        

130,  245 CaC2          

6,  12,  76 Ca(OH)2     77,  84,  130,  169 

. CaCN2        245 

CaSO4        12,  399 

69,  121 CaCO3        74,  75,  79 


122  APPENDIX 


124...  

CaSiO3 

,  274,  275 

273,  274  

Ca(PO3)2 

275  

Ca3(P04)2 

273 

CARBON 

244,  246,  247,  258,  273,  276, 

C 

245 

420 

C2H2 

130 

234,400  

CH4 

131 

167  

CC14 

192,242  

CO 

246,  244,  258,  273,  274,  275 

71,  79  

CO2 

20,    21,    69,    134,    242,    288, 

400,  401,  420,  430,  431 

127,  401  

CS2 

145-7,  334                 ... 

CN- 

C2N2 

254 

CH3C1 

...  234 

288  

CH2O 

COC12 

167 

COC1 

276 

CHLORINE 

27,   53,   177,   182,   183,   214, 

C12 

196,  198,  200,  201,  215,  218, 

234,  276,  280,  292 

219,  251,  253,  371,  410,  411, 

415,  437,  443 

387,  391  

ci- 

27,  53 

2,  4,  6,  8,  19,  110,  118,  196, 

HC1 

101,  119,  234,  235,  283,  284, 

223,  287,  371 

372,  421,  443 

262  

C1O2 

HOC1 

235,  282 

CHROMIUM 

298,  305 

Cr++ 

Qr+++ 

35,  306,  310,  312,  317 

303  

CrOr 

;  302 

140,  382,  433  

CrO4~ 

25,  303,  305 

25    35    281    306    310    312 

Cr207— 

317,  372 

CrCl3 

.      .  ;    253 

Cr2O3 

201,  252,  268  281 

252  253 

CrO3 

301,  302,  432  

Cr(OH)3 

298 

201 

Cr(OCl)2 

cwsa)a 

..432 

APPENDIX  123 

COBALT 

Co           430 

142,  297 Co++          

CoO          431 

191 CoO3          

285 Co(OH)2      142 

255 Co(OH)3      285,  297 

331 Co(NO2)2      

100 Co(NO2)3      331 

CoSo4         255 

431 CoCO3        

430 Co(OOC)2 

MgCoO3       * 191 

COPPER 

51,  207,  324,  328,  330 Cu  . .  26,  50,  56,  208,  229,  240,241 

26,  50,  56,  90,  141,  148,  153.  Cu++         51,  324,  329,  330 

134 CuCl          

135,  208 ( CuCl2 19,  207 

Cu2O         ..134 

19,  229,  240,  329 CuO          • 103,  104,  328 

81,  240,  241 CuS          153 

103,  149 Cu(OH)2 141,  159 

CuCN        254 

104 Cu(NO3)2      

1086,  159,  241,  254 CuSO4          108a 

CuCO3        135 

Cu(NH3)4++   90,  149 

Cu3SbS3       81 

Cu2Fe(CN6)     . , 148 

FLUORINE 

181,  215,  216,  217,  415,  418. .  F2            

124 HF  .,  .  .181,  216,  217,  415,  418 

IF5           236 

GALLIUM 

193 Ga           

GaCl2 193 

193 GaCl3         

, Ga2O3         422 

422 Ga(NO3)3      


124 


APPENDIX 


GOLD 

266  

Au 

367,  370 

AuCl 

437 

367,  368  

AuCl3 

438 

137,  370  

Au2O3 

369 

369  

Au(OH)3 

368 

437,  438  

HAuCl4 

266 

HYDROGEN 

40,    44,    176,    181,    228-232, 

H2 

42,  43,  41,  222-227,  416,  442 

277,  281,  293,  384 

16,  33,  34,  35,  45,  306  to  315, 

H+ 

23,  332 

323  to  33  f 

16,  24,  23,  46,  295  to  298,  302, 

OH- 

304-8,  440 

20,21,41-3,216,335,338... 

H2O 

.  .4-15,  82-86,  218,  300-331 

295-309,  363,  367,  370,  373.  . 

H2O2 

407 

IODINE 

55,  184,  236,  286,  414  

li 

54,  199,  213,  287,  299,  312- 

316,  376-379,  419 

54,  434  

I- 

55 

199,  312-316  

HI 

286,  375,  379,  414 

IRON 

26,  41,  42,  43,  56,  178,  179, 

Fe 

40,  44,  205,  209 

195,  206,  208,  212 

34,  145,  296,  317-319  

Fe++ 

....26,  56 

22                ... 

Fe+++ 

34,  317-319 

209  

FeCl2 

117,  208 

3              

FeCl3 

8 

FeBra 

383 

40,  44,  205  

Fe2O3 

41,  42,  43,  179,  206 

117,  187  

FeS 

178,  212 

8,  14,  383  

Fe(OH)3 

279,296 

FeSO4 

187,  195 

195 

Fe-zfSOJi 

14 

Fe2(CO3)3 

22 

97  

Fe(CN)2 

145 

Fe(CNS)3 

3 

148 

Fe(GN)6~~ 

280 

K4Fe(CN)6 

97 

Fe(CN)8~~ 

K3Fe(CN)6 

.  .280 

APPENDIX  125 

LEAD 

50,  59,  423 Pb            49,  65,  239 

23,  49,  61,  65,  433,  434 Pb++          50,  59,  64,  327 

357,  359 PbCl2        410 

435,  410 PbCl4         408 

PbI2          .  1,  434 

239 PbO           204,  237,  423 

60,  267 PbO2 435 

62,  68,  408 PbO2~         60,  349 

80,  237,  239,  327 PbS          22,  358,  359 

1,  204,  346,  349 Pb(NO3)2      267 

64,  66 PbSO4     

348 Pb(OH)2      .346,  347 

PbCr04       433 

.  PbSb2S4 80 

MAGNESIUM 

224,  227,  247 Mg 

174,  219 MgCl2         397 

191 MgO          174,  219,  227,  247 

Mg(N03)2     224 

MgCoO3       191 

MANGANESE 

295,  304,  387 Mn++        33,  34,  308,  309,  311,    313, 

318 

34,   265,  307,  309,  311,  313  MnOr 

318 

MnO4~     259 

373,  385 MnCl2.       371,  387,  411 

411 MnCl4        409 

' MnO           293 

32,   33,   259,   293,  308,   371,  MnO2         300,  304,  307 

392,  409 

366 Mn(OH)2      365 

203,  293,  300 Mn(OH)3      295 

HMriO2          203 

365 HMnO4         364 

H2MnO3        373 

MnSO4            32,  265,  363,  366,  392 

MnCO3                                       385 


MERCURY 

52,  194,  325,  389 Hg  28,   51,   197,   207,   269,   319, 

332,  334 


126 


APPENDIX 


332,  334,  382 

51,  319,  389 

333 

207,  233,  270,  354,  356. 

269 

28,  197 


Hg+ 


...339 
52,  325 
..270 


194 

268.. 


HgCl  

HgCl2  

Hg2O  „ 268 

HgO 

HgS  332,  355,  356 

HgNO3  194 

Hg(N03)2  

Hg2CrO4  382 

Hg(CN)2  334 

Hg(NH2)Cl  233,333 


136. 


MOLYBDENUM 
Mo03 

NICKEL 


.294 


89.... 

Ni++ 

99.... 

Ni(CN)2 

NUNHjOe*4- 

.      .89 

K2Ni(CN)4 

99 

180,  245...,  

NITROGEN 

N2 

248,  252,  287 

287 

N2H4 

NH3 

(see  ammonia)        .             .  . 

189,  190,  265 

N2O2 

266,  323,  325,  327,  331 

N2O3 

30 

NO2 

180,  189,  204,  261,  288,  289, 

82 

N2O6 

290 
396,  422 

290 

H2N2O2 

291 

291   . 

HNO2 

5,  7,  9,  30  

HNO3 

82,  190 

35,  311,  310  

323,327  

NOr 

35,  310,  311 

282 

NC13 

291 

NH2OH 

29,  176,  179,  180,  187,  191, 

OXYGEN 
02 

28,  47,   197,   201,   202,    205, 

206,  218,  219,  220,  237,  249, 
259,  294 

419.. 


03 


217,  255,  257,  268,  277,  279, 
306-9,    363,    366,    367,    370, 
386,  393,  400,  401,  419 
216,  418 


APPENDIX  127 


PHOSPHORUS 

182,  263,  271.....  

P            

273,  274,  275 

PH3          

132,  271 

163  

PC13          

182,  251 

164,  251,  87,  133 

PC15 

171   220 

PBr3 

172 

PBr5 

87,  128  

P205          

220 

HPO3         

128,  263 

H3P03        

.  .  .  .  163,  171 

11,  13,  15  

H3P04        

164,  173 

POC1         

87 

173  

POC13        

133 

POBr3         

172 

132  

PHJ          

PLATINUM 

Pt           

200,384 

384  

PtS                .  .  .    . 

152  

ptci6—     

POTASSIUM 

192,  210,  248  

K            

226,  227 

152 

K+ 

KC1          

....3,  210,  256,  257 

1,  299,  375-9,  419... 

.  .    .  .            KI               

413. 

KBr 

78,  259  

K2O          

248 

5,  11  226,  227  

KOH         

299 

,  

KSH         

398 

97,  98,  99  

KCN         

254 

248,  250  

KNO3        

1,5 

100  

KN02        

250,  261 

K2SO4        

78,375-379 



K3PO4            

.11 

256.. 

KC1O 

KC102        

262 

257,  413  

KC103        

256,  262 

KC1O4        

257 

KBr03        ........ 

413 

397  

K2MgCl3      

279  

.  .    .  .        KoFeO4           

281,372  

K2Cr2O7       

K2Mn04      

259 

130  APPENDIX 

TIN 

30,326 Sn  

119,  129,  350,  353 SnCl2         

95 ...-.  SnS  119 

96 SnS2          

352 Sn(OH)2       129,  350  351 

H2SnO3 107,  326 

107 H4SnO4        30 

TELLURIUM 

436 Te  260 

TeH2          116 

TeO          436 

TeO3 105 

105 Te(OH)6 

260 H2TeCl6       

THALLIUM 

391 ..  Tl*  

292 T1C1          391,  392 

,V T1C13 292 

392 -,..         T1NO3 

THORIUM 

ThOs         .  .  . 396 

396 Th(N03)4      

TITANIUM 

276 TiO2 

395 TiCl4         ......  ? 276 

H4TiO4        395 

TUNGSTEN 

183 W  232 

170 WC16          183 

125,  232 WO3          

WO(OH)4 126,  170 

ZINC 
49,  185,  186,  209 

222,  225,  323  Zn          198,  210 

143,  147 Zn*4-          49,  323,  428,  429 

198,  210. . .. ZnCl, .211 


APPENDIX 


131 


188. 


ZnS 


..........................  Zn3Sb2 

24,  92,  144  ................  Zn(OH)2 

..........................  ZnO2— 

98  ........  .  ...............  Zn(CN), 

427,  428  ...................  ZnSO4 


186 

185 

143 

.  24,  144,  225 

147 

188,  222 


K2Zn(CN)4    4, 98 

Zn(NH3)6Cl    92 


134  INDEX   AND   GLOSSARY 

CLASSIFICATION  OF  ELEMENTS  is  derived  from  the  periodic 

system 109 

COLLOIDS  are  particles  of  ultra-microscopic  size 51 

COLOR  often  indicates  the  stage  of  oxidation 106 

— change  indicates  a  chemical  reaction 23 

COMBINATION  is  a  chemical  reaction  in  which  two  elements  unite 
and  form  a  compound,  one  element  being  reduced,  the  other 

oxidized 87 

COMPLEX-ION  is  a  radicle  or  group  of  atoms  which  is  electrically 

charged,  either  positively  or  negatively 3 

COMPLEX  REACTION  results  from  chemical  changes  in  which 

two  or  more  types  of  reaction  simultaneously  take  place 96 

COMPOUNDS  are  molecules  with  unlike  atoms 6 

— the  sum  of  valence  numbers  in  stable  compounds  is  zero 14 

— the  weight  relation  of  their  constituents  is  shown  in  a  formula ...  6 

— the  percentage  composition  is  calculated  from  formula 6,  19 

— the  same  element  may  form  different  series  of  compounds 13 

- — the  names  of  different  series  depend  upon  valence  number 17,  103 

CONCENTRATION  is  the  amount  of  atoms  or  molecules  in  a  cer- 
tain volume;  cone,  of  gases  is  expressed  in  terms  of  pressure; 

cone,  of  substances  in  solutions  is  expressed  in  moles  per  liter  ...  53 

— change  in  concentration  affects  the  chemical  equilibrium 56 

DECOMPOSITION  is  the  reverse  reaction  of  addition 82 

DISPLACEMENT  is  a  chemical  reaction  in  which  one  element 
exchanges  charges  with  another  element — it  is  either  oxidation 

or  reduction 69,  89 

DISPLACEMENT  SERIES  is  an  arrangement  of  the  elements  in 

the  order  in  which  they  hold  their  electric  charges 70,  107 

— indicates  the  electro  motive  force  of  elements 68 

DISSOCIATION  is  the  breaking  apart  of  molecules 63 

— thermal  dissociation  takes  place  under  the  influence  of  heat    ....     64 
— electrical  dissociation  or  ionization  is  produced  by  electricity    ...  3,  65 
DIVISION  is  the  reverse  reaction  of  combination  in  which  a  com- 
pound breaks  apart,   one  element  being  oxidized,   the  other 
reduced 88 

E.  M.  F.  =  electro-motive  force 68 

EARTH  ALKALI  METALS  are  the  elements  of  the  second  group 

of  the  periodic  system:  Mg,  Ca,  Sr,  Ba Ill 

— they  are  bivalent  but  univalent 13,  110 

EARTH  METALS  are  the  elements  of  the  third  group  of  the  periodic 

system  e.g.,  Al,  Sc Ill 

— they  are  trivalent,  but  univalent 13,  110 


INDEX   AND   GLOSSARY  135 

ELEMENTS  are  chemically  indivisable  and  indestruc table  sub- 
stances whose  molecules  consist  of  like  atoms 2 

— in  their  free  state  they  have  a  valence  number  of  zero* 12 

— those  having  in  their  compounds  but  one  valence  number  are 

univalent,  those  with  two  or  more  are  polyvalent 13 

ELECTRIC  BATTERY  is  formed  when  two  elments  are  connected 
by  a  conducting  media  but  are  locally  separated  so  that  a  dis- 
placement reaction  can  take  place 71 

ELECTRIC  DISSOCIATION  is  ionization 3,  65 

ELECTRO  AFFINITY  or  e.m.f.  is  the  force  with  which  the  atoms 

hold  their  equally  large  ionic  charges 68 

ELECTRODE  is  the  positive  or  negative  pole 65,  71 

ELECTROLYSIS  is  the  spacial  or  local  separation  of  ions  (cations 

and  anions)  by  an  electric  current 65 

ELECTRO-MOTIVE    FORCE    is    the    power    of    an    atom     to 

retain  electrical  charges  as  expressed  in  the  displacement  series     68 
ELECTRO-POTENTIAL  is  the  tension  between  two  different  sets 

of  atoms  and  their  ions,  expressed  in  volts 71 

ELEMENT  is  an  aggregation  of  matter  consisting  of  one  type  of 

atoms 2 

ENDOTHERMIC  COMPOUND  requires  heat  for  its  formation, 

,  and  liberates  heat  in  its  decomposition 62 

ENDOTHERMIC    REACTION    always    proceeds    slowly    and 

requires  heat  which  it  absorbs 62 

EQUATION  is  the  expression  of  a  chemical  reaction 23 

— there  are  non-ionic  or  molecular  and  ionic  equations 28 

— balancing  of  equations <•     32 

— finishing  of  incomplete  equations 29 

— the  universal  equation  of  neutralization  is  H+  -f  OH~  =  H2O    .  .     27 
EQUILIBRIUM  is  the  balanced  state  reached  in  a  chemical  reaction 
when  the  concentration  between  the  reacting  substances   has 
become  such,  that  decomposition  and  re-combination  proceeds 

with  equal  speed 55 

— the  removal  of  one  reaction-product  shifts  the  equilibrium  and  the 

reaction  may  proceed  to  completion 56 

EXOTHERMIC  COMPOUND  liberates  heat  in  its  formation,  and 

requires  or  absorbs  heat  in  its  decomposition 62 

EXOTHERMIC  REACTION  proceeds  rapidly  and  liberates  heat.     62 
EXPLOSIONS  are  exothermic  reactions 62 

FORMULA  is  a  combination  of  symbols  showing  the  composition  of 

a  molecule 6,  10 

— There  are  four  types  of  formula:  empirical,  constitutional, 

rational,  and  structural 8 


136  INDEX   AND   GLOSSARY 

— Summary  of  the  information  contained  in  formula: 18 

GAS  formation  indicates  a  chemical  reaction 23 

— density,  weight  of  1  liter,  specific  gravity,  and  volume  relation 

of  a  gas  is  shown  in  the  formula 19 

GOLD  PERIOD  is  composed  of  the  elements  of  the  fifth  sub-period 

of  the  periodic  system Ill 

GRAM  MOLECULE  is  the  molecular  weight  of  a  compound  in 

grams 19 

HEAT  increases  the  speed  of  reactions 61 

— produces  thermal  dissociation, 62 

— is  absorbed  in  endothermic  reactions  and  liberated  in  exothermic 

reactions, : 63 

— is  absorbed  in  the  formation  of  endothermic  compounds  and 

liberated  in  the  formation  of  exothermic  compounds, 62 

HEAVY  METALS  are  the  elements  of  the  sub-periods  of  the 

periodic  system  (lower  part  of  chart) 110,  113 

— are  all  polyvalent,  therefore  have  different  series  of  compounds 

(-ous  and  -ic  compounds) 13,  17 

— may  be  amphoteric,  and  form  many  complex  compounds 113 

HYDROGEN  is  the  basis  of  valency  and  valence  numbers 11 

— Elements  combining  with  H  have  a  negative  valence  number. ...  12 

— Elements  replacing  H  have  a  positive  valence  number 12 

HYDROLYSIS  is  the  reverse  reaction  of  neutralisation 86 

INERT  ELEMENTS   (or  noble  gases)  do  not  combine  with  any 

other  element Ill,  113 

IONIC  REACTIONS  occur  only  when  the  substance  is  in  solution     57 
— proceed  very  rapidly,  and  depend  upon  the  formation  of  one  of 

7  compounds 57 

IONIZATION  or  electric  dissociation  is  the  breaking  apart  of 
molecules  (electrically  neutral)  into  electrified  atoms  (charged 
positively  or  negatively)  taking  place  when  certain  substances 

are  dissolved  in  water 3 

IONS  are  electrically  charged  atoms  or  groups  of  atoms 3 

— They  may  have  one  or  more  positive  (cations)  or  negative  (anions) 

charges 4 

— Ionic  charges  are  balanced  before  the  atoms  are  balanced. 31 

— and  after  oxidation  and  reduction  is  balanced 41 

IRON-PERIOD,  the  elements  of  the  third  sub-period  of  the  periodic 

system  Cr,  Mn,  Fe,  Co,  Ni,  Cu Ill,  113 

LIGHT  METALS  are  the  elements  of  the  a-sub-periods  of  the 

periodic  system Ill,  113 


INDEX   AND   GLOSSARY  137 

METATHESIS  is  a  chemical  reaction  with  an  even  exchange 83 

MOLAR  SOLUTIONS  contain  the  gram  molecular  weight  of  sub- 
stance in  1  liter 53 

MOLE  is  the  molecular  weight  of  an  element  or  compound  in  grams 

( =  gram  molecule) 19,  53 

MOLECULE  is  the  chemical  combination  of  two  or  more  like  or 

unlike  atoms 2 

MOLECULAR  WEIGHT  is  the  relative  mass  of  a  molecule  which 

is  found  by  adding  the  atomic  weights 6 

NEUTRALIZATION  is  a  metathesical  reaction  between  an  acid  and 
a  base  which  gives  a  salt  and  water — the  universal  equation 

is  H+  +  OH-  =  H2O 27,  85 

NOBLE  GASES  are  the  inert  elements  of  the  zero  group Ill 

NOBLE  METALS  are  the  elements  of  group  8  and  Ib  of  the  silver- 

and  gold  period Ill,  113 

NOMENCLATURE  of  compounds  depends  mainly  on  the  valence 

number , 17 

— table  and  general  rules 103 

NON-METALS  are  the  elements  of  the  b-sub-periods  of  the  periodic 

system : Ill 

— are  strongly  electro-negative,  form  acids,  are  polyvalent 113 

NORMAL  SOLUTIONS  contain  the  gram  equivalent  weight  of  the 

substance  in  1  liter  solution 53 

OXIDATION  is  the  increase  or  augmentation  of  the  valence 
number 15,  37 

POLAR  NUMBER  is  the  same  as  valence  number 12 

PERCENTAGE  of  constituents  in  a  compound  is  calculated  from 

the  formula 6 

PERIODIC  SYSTEM  is  a  logical  and  natural  classification  of  all 

elements  according  to  their  properties 109 

— aids  to  memorize  the  properties  of  elements,  and  enables  the  pre- 
diction of  properties  of  unknown  elements .  •  •  •  •  HO 

PHOSPHOR  GROUP  are  the  elements  of  the  fifth  group  of  the 

periodic  system Ill 

PRECIPITATE  formation  is  due  to  an  insoluble  compound  and 

indicates  a  chemical  reaction 23 

PREPARATION  OF  SALTS  is  tabulated  on  page 118 

RADICLE  is  a  group  of  atoms  having  one  or  more  free  valencies. 

It  reacts  like  an  atom   and  cannot  exist  in  the  free  state.     26 


138  INDEX   AND   GLOSSARY 

RADIOACTIVE  ELEMENTS  are  the  elements  of  highest  atomic 

weight  which  spontaneously  disintegrate  or  break  apart Ill 

RARE  EARTH  METALS  are  the  elements  of  the  V"  sub-period.  .    Ill 

REACTION  is  a  chemical  change  or  interaction  of  molecules 23 

— Control  of  reaction  depends  on  concentration,  velocity,  and  the 

e.m.f 48,  75 

— Control  of  ionic  reactions  depends  on  the  formation  of  one  of 

seven  substances 57,  75 

REDUCTION  is  the  decrease  or  diminution  of  the  valence  number  15,  37 
RESTITUTION  is  the  reverse  reaction  of  substitution 91 

SALTS  are  generally  formed  from  acids  and  bases  (neutralization)  26 
— They  may  be  prepared  from  metal,  metaloxides,  metalhydroxides, 

metalcarbonates,  metal  sulfides 118 

SILVER  PERIOD  are  the  elements  of  the  fourth  sub-period.  . 111 

SOLUBILITY  TABLE 115 

SUBSTITUTION  is  a  chemical  reaction  in  which  an  element  is 

either  oxidised  or  reduced  and  combines  with  the  displaced 

element 91 

SULFUR  GROUP  are  the  elements  of  the  sixth  group  of  the  periodic 

system:  O,  S,  Se,  Te Ill 

SYMBOL  is  one  or  two  letters  representing  an  atom  and  its  relative 

mass 1 

— is  used  to  represent  an  atom,  molecule,  or  ion 4 

VALENCE  NUMBER  is  the  arithmetical  expression  of  valency. 

It  can  be  either  +  or  — ,  and  is  based  upon  H  =  1,  andO  =  —2.12,  14 
— of  the  same  element  may  be  different  and  then  it  indicates  different 

series  of  compounds 13 

— of  free  elements  is  always  zero 12 

— the  sum  in  a  stable  compound  is  always  zero 14 

VALENCY  is  the  capacity  of  an  atom  to  combine  with  other  atoms 

in  a  definite  proportion 11 

— is  measured  with  regard  to  H  and  can  be  any  integer  from  1—8  12 

— when  unknown  may  be  calculated  from  the  formula 14 

WATER  formation  is  the  characteristic  earmark  of  neutralization  .  27 
WATER  OF  CRYSTALLIZATION  is  the  water  contained  in 

crystals 7 


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